141149
An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s) - velocity (v) graph of this object is
1
2
3
4
Explanation:
C From Newton's third equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\) \(\mathrm{v}^{2}=2 \mathrm{as} \quad[\because \mathrm{u}=0]\) We know, Parabola equation, \(x^{2}=2\) ay
AP EAMCET-20.08.2021
Motion in One Dimensions
141150
The relation between time \(t\) and distance \(x\) is \(t\) \(=a x^{2}+b x\), where \(a\) and \(b\) are constants. The acceleration is
1 \(-2 a b v^{2}\)
2 \(2 \mathrm{bv}^{3}\)
3 \(-2 \mathrm{av}^{3}\)
4 \(2 \mathrm{av}^{2}\)
Explanation:
C Let the acceleration be A. \(\Rightarrow \quad \mathrm{t}=\mathrm{ax}^{2}+\mathrm{bx}\) On differentiating the above equation with respect to \(t\). Then, \(\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{ax}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{bx})\) \(1=2 \mathrm{ax} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{b} \frac{\mathrm{dx}}{\mathrm{dt}}\) \(1=\frac{\mathrm{dx}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Again differentiating the equation number (ii) w.r.t. \(t\) \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{(2 \mathrm{ax}+\mathrm{b}) \frac{\mathrm{d}}{\mathrm{dt}}(1)-1 \cdot \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\because \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{a} \frac{\mathrm{dx}}{\mathrm{dt}}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\therefore \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{av}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) From equation no. (ii) \(\mathrm{v}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Acceleration \(=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=-2 \mathrm{av} \times \frac{1}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) Acceleration \(=-2 \mathrm{av} \times \mathrm{v}^{2}\) Acceleration \(=-2 \mathrm{av}^{3}\)
MP PET-2012
Motion in One Dimensions
141151
A car moving with a speed of \(40 \mathrm{~km} / \mathrm{h}\) can be stopped after \(2 \mathrm{~m}\) by applying brakes. If the same car is moving with a speed of \(80 \mathrm{~km} / \mathrm{h}\), what is the minimum stopping distance?
1 \(8 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
A For first case : \(\mathrm{u}=40 \mathrm{~km} / \mathrm{h}\) \(=\left(40 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) For third equation of motion we have, \(v^{2}=u^{2}+2 a s\) \(0=\left(40 \times \frac{5}{18}\right)^{2}+2 a(2)\) \(0=\left(\frac{200}{18}\right)^{2}+4 a\) \(\mathrm{a}=-\frac{1}{4}\left(\frac{100}{9}\right)^{2}\) For second case :- \(\mathrm{u}=80 \mathrm{~km} / \mathrm{h}\) \(\mathrm{u}=\left(80 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(0=\left(80 \times \frac{5}{18}\right)^{2}+2 \times \mathrm{a} \times \mathrm{s}\) \(2 \mathrm{as}=-\left(\frac{200}{9}\right)^{2}\) Putting the value of ' \(a\) ' in equation (ii) \(-2 \times \frac{1}{4} \times\left(\frac{100}{9}\right)^{2} \times s=-\left(\frac{200}{9}\right)^{2}\) \(\frac{2}{4} \times \frac{100}{9} \times \frac{100}{9} \times s=\left(\frac{200}{9}\right)^{2}\) \(\frac{s}{2}=2 \times 2\) \(s=2 \times 2 \times 2\) Distance, \(\mathrm{s}=8 \mathrm{~m}\)
AIEEE-2003
Motion in One Dimensions
141152
A particle moves along a straight line \(O X\). At a time \(t\) (in seconds) the distance \(x\) (in meters) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\) How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given that, Distance \(\mathrm{x}=40+12 \mathrm{t}-\mathrm{t}^{3}\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\mathrm{v} =\frac{\mathrm{d}}{\mathrm{dt}}\left(40+12 \mathrm{t}-\mathrm{t}^{3}\right)\) \(\mathrm{v} =12-3 \mathrm{t}^{2}\) Final velocity of the particle is zero \((\mathrm{v}=0)\) So, \(\quad 12-3 \mathrm{t}^{2}=0\) \(\mathrm{t}^{2}=\frac{12}{3}\) \(\mathrm{t}=\sqrt{4}\) \(\mathrm{t}=2 \mathrm{sec} .\) Distance covered by the particle before coming to rest is \(\mathrm{x} =40+12 \mathrm{t}-\mathrm{t}^{3}\) \(=40+12 \times 2-2^{3}\) \(=64-8\) \(=56 \mathrm{~m}\)
WE JEE-2012
Motion in One Dimensions
141153
A particle starts its motion from rest under the action of a constant force. If the distance covered in first \(10 \mathrm{~s}\) is \(S_{1}\) and that covered in the first \(20 \mathrm{~s}\) is \(S_{2}\), then
1 \(s_{2}=2 s_{1}\)
2 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
Explanation:
C Particle starts motion from rest, \(\mathrm{u}=0\) Let, acceleration \(=\mathrm{a} \mathrm{m} / \mathrm{s}^{2}\), time \(\mathrm{t}_{1}=10 \mathrm{sec}\) velocity after 10 second \(=\mathrm{v}_{1} \mathrm{~m} / \mathrm{s}\) So, \(\quad \mathrm{v}_{1}=\mathrm{u}+\mathrm{at}_{1}\) \(\mathrm{v}_{1}=0+\mathrm{a} \times 10\) \(\mathrm{v}_{1}=10 \mathrm{a} \mathrm{m} / \mathrm{s}\) And \(\quad \mathrm{v}_{1}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{1}\) \((10 \mathrm{a})^{2}=0+2 \mathrm{as}_{1}\) \(\mathrm{s}_{1}=\frac{100 \mathrm{a}^{2}}{2 \mathrm{a}}=50 \mathrm{a}\) Velocity after \(\mathrm{t}_{2}=20\) second \(\mathrm{v}_{2}=\mathrm{u}+\mathrm{at}_{2}\) \(\mathrm{v}_{2}=0+\mathrm{a} \times 20\) \(\mathrm{v}_{2}=20 \mathrm{a}\) And \(\quad \mathrm{v}_{2}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{2}\) \((20 a)^{2}=0+2 a_{2}\) \(s_{2}=s_{1}=\frac{100 a^{2}}{2 a}=50 a\) So, \(\quad \frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{50 \mathrm{a}}{200 \mathrm{a}}\) \(\frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{1}{4}\) \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
141149
An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s) - velocity (v) graph of this object is
1
2
3
4
Explanation:
C From Newton's third equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\) \(\mathrm{v}^{2}=2 \mathrm{as} \quad[\because \mathrm{u}=0]\) We know, Parabola equation, \(x^{2}=2\) ay
AP EAMCET-20.08.2021
Motion in One Dimensions
141150
The relation between time \(t\) and distance \(x\) is \(t\) \(=a x^{2}+b x\), where \(a\) and \(b\) are constants. The acceleration is
1 \(-2 a b v^{2}\)
2 \(2 \mathrm{bv}^{3}\)
3 \(-2 \mathrm{av}^{3}\)
4 \(2 \mathrm{av}^{2}\)
Explanation:
C Let the acceleration be A. \(\Rightarrow \quad \mathrm{t}=\mathrm{ax}^{2}+\mathrm{bx}\) On differentiating the above equation with respect to \(t\). Then, \(\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{ax}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{bx})\) \(1=2 \mathrm{ax} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{b} \frac{\mathrm{dx}}{\mathrm{dt}}\) \(1=\frac{\mathrm{dx}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Again differentiating the equation number (ii) w.r.t. \(t\) \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{(2 \mathrm{ax}+\mathrm{b}) \frac{\mathrm{d}}{\mathrm{dt}}(1)-1 \cdot \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\because \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{a} \frac{\mathrm{dx}}{\mathrm{dt}}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\therefore \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{av}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) From equation no. (ii) \(\mathrm{v}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Acceleration \(=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=-2 \mathrm{av} \times \frac{1}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) Acceleration \(=-2 \mathrm{av} \times \mathrm{v}^{2}\) Acceleration \(=-2 \mathrm{av}^{3}\)
MP PET-2012
Motion in One Dimensions
141151
A car moving with a speed of \(40 \mathrm{~km} / \mathrm{h}\) can be stopped after \(2 \mathrm{~m}\) by applying brakes. If the same car is moving with a speed of \(80 \mathrm{~km} / \mathrm{h}\), what is the minimum stopping distance?
1 \(8 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
A For first case : \(\mathrm{u}=40 \mathrm{~km} / \mathrm{h}\) \(=\left(40 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) For third equation of motion we have, \(v^{2}=u^{2}+2 a s\) \(0=\left(40 \times \frac{5}{18}\right)^{2}+2 a(2)\) \(0=\left(\frac{200}{18}\right)^{2}+4 a\) \(\mathrm{a}=-\frac{1}{4}\left(\frac{100}{9}\right)^{2}\) For second case :- \(\mathrm{u}=80 \mathrm{~km} / \mathrm{h}\) \(\mathrm{u}=\left(80 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(0=\left(80 \times \frac{5}{18}\right)^{2}+2 \times \mathrm{a} \times \mathrm{s}\) \(2 \mathrm{as}=-\left(\frac{200}{9}\right)^{2}\) Putting the value of ' \(a\) ' in equation (ii) \(-2 \times \frac{1}{4} \times\left(\frac{100}{9}\right)^{2} \times s=-\left(\frac{200}{9}\right)^{2}\) \(\frac{2}{4} \times \frac{100}{9} \times \frac{100}{9} \times s=\left(\frac{200}{9}\right)^{2}\) \(\frac{s}{2}=2 \times 2\) \(s=2 \times 2 \times 2\) Distance, \(\mathrm{s}=8 \mathrm{~m}\)
AIEEE-2003
Motion in One Dimensions
141152
A particle moves along a straight line \(O X\). At a time \(t\) (in seconds) the distance \(x\) (in meters) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\) How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given that, Distance \(\mathrm{x}=40+12 \mathrm{t}-\mathrm{t}^{3}\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\mathrm{v} =\frac{\mathrm{d}}{\mathrm{dt}}\left(40+12 \mathrm{t}-\mathrm{t}^{3}\right)\) \(\mathrm{v} =12-3 \mathrm{t}^{2}\) Final velocity of the particle is zero \((\mathrm{v}=0)\) So, \(\quad 12-3 \mathrm{t}^{2}=0\) \(\mathrm{t}^{2}=\frac{12}{3}\) \(\mathrm{t}=\sqrt{4}\) \(\mathrm{t}=2 \mathrm{sec} .\) Distance covered by the particle before coming to rest is \(\mathrm{x} =40+12 \mathrm{t}-\mathrm{t}^{3}\) \(=40+12 \times 2-2^{3}\) \(=64-8\) \(=56 \mathrm{~m}\)
WE JEE-2012
Motion in One Dimensions
141153
A particle starts its motion from rest under the action of a constant force. If the distance covered in first \(10 \mathrm{~s}\) is \(S_{1}\) and that covered in the first \(20 \mathrm{~s}\) is \(S_{2}\), then
1 \(s_{2}=2 s_{1}\)
2 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
Explanation:
C Particle starts motion from rest, \(\mathrm{u}=0\) Let, acceleration \(=\mathrm{a} \mathrm{m} / \mathrm{s}^{2}\), time \(\mathrm{t}_{1}=10 \mathrm{sec}\) velocity after 10 second \(=\mathrm{v}_{1} \mathrm{~m} / \mathrm{s}\) So, \(\quad \mathrm{v}_{1}=\mathrm{u}+\mathrm{at}_{1}\) \(\mathrm{v}_{1}=0+\mathrm{a} \times 10\) \(\mathrm{v}_{1}=10 \mathrm{a} \mathrm{m} / \mathrm{s}\) And \(\quad \mathrm{v}_{1}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{1}\) \((10 \mathrm{a})^{2}=0+2 \mathrm{as}_{1}\) \(\mathrm{s}_{1}=\frac{100 \mathrm{a}^{2}}{2 \mathrm{a}}=50 \mathrm{a}\) Velocity after \(\mathrm{t}_{2}=20\) second \(\mathrm{v}_{2}=\mathrm{u}+\mathrm{at}_{2}\) \(\mathrm{v}_{2}=0+\mathrm{a} \times 20\) \(\mathrm{v}_{2}=20 \mathrm{a}\) And \(\quad \mathrm{v}_{2}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{2}\) \((20 a)^{2}=0+2 a_{2}\) \(s_{2}=s_{1}=\frac{100 a^{2}}{2 a}=50 a\) So, \(\quad \frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{50 \mathrm{a}}{200 \mathrm{a}}\) \(\frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{1}{4}\) \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
141149
An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s) - velocity (v) graph of this object is
1
2
3
4
Explanation:
C From Newton's third equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\) \(\mathrm{v}^{2}=2 \mathrm{as} \quad[\because \mathrm{u}=0]\) We know, Parabola equation, \(x^{2}=2\) ay
AP EAMCET-20.08.2021
Motion in One Dimensions
141150
The relation between time \(t\) and distance \(x\) is \(t\) \(=a x^{2}+b x\), where \(a\) and \(b\) are constants. The acceleration is
1 \(-2 a b v^{2}\)
2 \(2 \mathrm{bv}^{3}\)
3 \(-2 \mathrm{av}^{3}\)
4 \(2 \mathrm{av}^{2}\)
Explanation:
C Let the acceleration be A. \(\Rightarrow \quad \mathrm{t}=\mathrm{ax}^{2}+\mathrm{bx}\) On differentiating the above equation with respect to \(t\). Then, \(\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{ax}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{bx})\) \(1=2 \mathrm{ax} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{b} \frac{\mathrm{dx}}{\mathrm{dt}}\) \(1=\frac{\mathrm{dx}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Again differentiating the equation number (ii) w.r.t. \(t\) \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{(2 \mathrm{ax}+\mathrm{b}) \frac{\mathrm{d}}{\mathrm{dt}}(1)-1 \cdot \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\because \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{a} \frac{\mathrm{dx}}{\mathrm{dt}}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\therefore \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{av}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) From equation no. (ii) \(\mathrm{v}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Acceleration \(=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=-2 \mathrm{av} \times \frac{1}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) Acceleration \(=-2 \mathrm{av} \times \mathrm{v}^{2}\) Acceleration \(=-2 \mathrm{av}^{3}\)
MP PET-2012
Motion in One Dimensions
141151
A car moving with a speed of \(40 \mathrm{~km} / \mathrm{h}\) can be stopped after \(2 \mathrm{~m}\) by applying brakes. If the same car is moving with a speed of \(80 \mathrm{~km} / \mathrm{h}\), what is the minimum stopping distance?
1 \(8 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
A For first case : \(\mathrm{u}=40 \mathrm{~km} / \mathrm{h}\) \(=\left(40 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) For third equation of motion we have, \(v^{2}=u^{2}+2 a s\) \(0=\left(40 \times \frac{5}{18}\right)^{2}+2 a(2)\) \(0=\left(\frac{200}{18}\right)^{2}+4 a\) \(\mathrm{a}=-\frac{1}{4}\left(\frac{100}{9}\right)^{2}\) For second case :- \(\mathrm{u}=80 \mathrm{~km} / \mathrm{h}\) \(\mathrm{u}=\left(80 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(0=\left(80 \times \frac{5}{18}\right)^{2}+2 \times \mathrm{a} \times \mathrm{s}\) \(2 \mathrm{as}=-\left(\frac{200}{9}\right)^{2}\) Putting the value of ' \(a\) ' in equation (ii) \(-2 \times \frac{1}{4} \times\left(\frac{100}{9}\right)^{2} \times s=-\left(\frac{200}{9}\right)^{2}\) \(\frac{2}{4} \times \frac{100}{9} \times \frac{100}{9} \times s=\left(\frac{200}{9}\right)^{2}\) \(\frac{s}{2}=2 \times 2\) \(s=2 \times 2 \times 2\) Distance, \(\mathrm{s}=8 \mathrm{~m}\)
AIEEE-2003
Motion in One Dimensions
141152
A particle moves along a straight line \(O X\). At a time \(t\) (in seconds) the distance \(x\) (in meters) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\) How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given that, Distance \(\mathrm{x}=40+12 \mathrm{t}-\mathrm{t}^{3}\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\mathrm{v} =\frac{\mathrm{d}}{\mathrm{dt}}\left(40+12 \mathrm{t}-\mathrm{t}^{3}\right)\) \(\mathrm{v} =12-3 \mathrm{t}^{2}\) Final velocity of the particle is zero \((\mathrm{v}=0)\) So, \(\quad 12-3 \mathrm{t}^{2}=0\) \(\mathrm{t}^{2}=\frac{12}{3}\) \(\mathrm{t}=\sqrt{4}\) \(\mathrm{t}=2 \mathrm{sec} .\) Distance covered by the particle before coming to rest is \(\mathrm{x} =40+12 \mathrm{t}-\mathrm{t}^{3}\) \(=40+12 \times 2-2^{3}\) \(=64-8\) \(=56 \mathrm{~m}\)
WE JEE-2012
Motion in One Dimensions
141153
A particle starts its motion from rest under the action of a constant force. If the distance covered in first \(10 \mathrm{~s}\) is \(S_{1}\) and that covered in the first \(20 \mathrm{~s}\) is \(S_{2}\), then
1 \(s_{2}=2 s_{1}\)
2 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
Explanation:
C Particle starts motion from rest, \(\mathrm{u}=0\) Let, acceleration \(=\mathrm{a} \mathrm{m} / \mathrm{s}^{2}\), time \(\mathrm{t}_{1}=10 \mathrm{sec}\) velocity after 10 second \(=\mathrm{v}_{1} \mathrm{~m} / \mathrm{s}\) So, \(\quad \mathrm{v}_{1}=\mathrm{u}+\mathrm{at}_{1}\) \(\mathrm{v}_{1}=0+\mathrm{a} \times 10\) \(\mathrm{v}_{1}=10 \mathrm{a} \mathrm{m} / \mathrm{s}\) And \(\quad \mathrm{v}_{1}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{1}\) \((10 \mathrm{a})^{2}=0+2 \mathrm{as}_{1}\) \(\mathrm{s}_{1}=\frac{100 \mathrm{a}^{2}}{2 \mathrm{a}}=50 \mathrm{a}\) Velocity after \(\mathrm{t}_{2}=20\) second \(\mathrm{v}_{2}=\mathrm{u}+\mathrm{at}_{2}\) \(\mathrm{v}_{2}=0+\mathrm{a} \times 20\) \(\mathrm{v}_{2}=20 \mathrm{a}\) And \(\quad \mathrm{v}_{2}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{2}\) \((20 a)^{2}=0+2 a_{2}\) \(s_{2}=s_{1}=\frac{100 a^{2}}{2 a}=50 a\) So, \(\quad \frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{50 \mathrm{a}}{200 \mathrm{a}}\) \(\frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{1}{4}\) \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in One Dimensions
141149
An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s) - velocity (v) graph of this object is
1
2
3
4
Explanation:
C From Newton's third equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\) \(\mathrm{v}^{2}=2 \mathrm{as} \quad[\because \mathrm{u}=0]\) We know, Parabola equation, \(x^{2}=2\) ay
AP EAMCET-20.08.2021
Motion in One Dimensions
141150
The relation between time \(t\) and distance \(x\) is \(t\) \(=a x^{2}+b x\), where \(a\) and \(b\) are constants. The acceleration is
1 \(-2 a b v^{2}\)
2 \(2 \mathrm{bv}^{3}\)
3 \(-2 \mathrm{av}^{3}\)
4 \(2 \mathrm{av}^{2}\)
Explanation:
C Let the acceleration be A. \(\Rightarrow \quad \mathrm{t}=\mathrm{ax}^{2}+\mathrm{bx}\) On differentiating the above equation with respect to \(t\). Then, \(\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{ax}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{bx})\) \(1=2 \mathrm{ax} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{b} \frac{\mathrm{dx}}{\mathrm{dt}}\) \(1=\frac{\mathrm{dx}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Again differentiating the equation number (ii) w.r.t. \(t\) \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{(2 \mathrm{ax}+\mathrm{b}) \frac{\mathrm{d}}{\mathrm{dt}}(1)-1 \cdot \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\because \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{a} \frac{\mathrm{dx}}{\mathrm{dt}}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\therefore \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{av}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) From equation no. (ii) \(\mathrm{v}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Acceleration \(=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=-2 \mathrm{av} \times \frac{1}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) Acceleration \(=-2 \mathrm{av} \times \mathrm{v}^{2}\) Acceleration \(=-2 \mathrm{av}^{3}\)
MP PET-2012
Motion in One Dimensions
141151
A car moving with a speed of \(40 \mathrm{~km} / \mathrm{h}\) can be stopped after \(2 \mathrm{~m}\) by applying brakes. If the same car is moving with a speed of \(80 \mathrm{~km} / \mathrm{h}\), what is the minimum stopping distance?
1 \(8 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
A For first case : \(\mathrm{u}=40 \mathrm{~km} / \mathrm{h}\) \(=\left(40 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) For third equation of motion we have, \(v^{2}=u^{2}+2 a s\) \(0=\left(40 \times \frac{5}{18}\right)^{2}+2 a(2)\) \(0=\left(\frac{200}{18}\right)^{2}+4 a\) \(\mathrm{a}=-\frac{1}{4}\left(\frac{100}{9}\right)^{2}\) For second case :- \(\mathrm{u}=80 \mathrm{~km} / \mathrm{h}\) \(\mathrm{u}=\left(80 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(0=\left(80 \times \frac{5}{18}\right)^{2}+2 \times \mathrm{a} \times \mathrm{s}\) \(2 \mathrm{as}=-\left(\frac{200}{9}\right)^{2}\) Putting the value of ' \(a\) ' in equation (ii) \(-2 \times \frac{1}{4} \times\left(\frac{100}{9}\right)^{2} \times s=-\left(\frac{200}{9}\right)^{2}\) \(\frac{2}{4} \times \frac{100}{9} \times \frac{100}{9} \times s=\left(\frac{200}{9}\right)^{2}\) \(\frac{s}{2}=2 \times 2\) \(s=2 \times 2 \times 2\) Distance, \(\mathrm{s}=8 \mathrm{~m}\)
AIEEE-2003
Motion in One Dimensions
141152
A particle moves along a straight line \(O X\). At a time \(t\) (in seconds) the distance \(x\) (in meters) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\) How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given that, Distance \(\mathrm{x}=40+12 \mathrm{t}-\mathrm{t}^{3}\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\mathrm{v} =\frac{\mathrm{d}}{\mathrm{dt}}\left(40+12 \mathrm{t}-\mathrm{t}^{3}\right)\) \(\mathrm{v} =12-3 \mathrm{t}^{2}\) Final velocity of the particle is zero \((\mathrm{v}=0)\) So, \(\quad 12-3 \mathrm{t}^{2}=0\) \(\mathrm{t}^{2}=\frac{12}{3}\) \(\mathrm{t}=\sqrt{4}\) \(\mathrm{t}=2 \mathrm{sec} .\) Distance covered by the particle before coming to rest is \(\mathrm{x} =40+12 \mathrm{t}-\mathrm{t}^{3}\) \(=40+12 \times 2-2^{3}\) \(=64-8\) \(=56 \mathrm{~m}\)
WE JEE-2012
Motion in One Dimensions
141153
A particle starts its motion from rest under the action of a constant force. If the distance covered in first \(10 \mathrm{~s}\) is \(S_{1}\) and that covered in the first \(20 \mathrm{~s}\) is \(S_{2}\), then
1 \(s_{2}=2 s_{1}\)
2 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
Explanation:
C Particle starts motion from rest, \(\mathrm{u}=0\) Let, acceleration \(=\mathrm{a} \mathrm{m} / \mathrm{s}^{2}\), time \(\mathrm{t}_{1}=10 \mathrm{sec}\) velocity after 10 second \(=\mathrm{v}_{1} \mathrm{~m} / \mathrm{s}\) So, \(\quad \mathrm{v}_{1}=\mathrm{u}+\mathrm{at}_{1}\) \(\mathrm{v}_{1}=0+\mathrm{a} \times 10\) \(\mathrm{v}_{1}=10 \mathrm{a} \mathrm{m} / \mathrm{s}\) And \(\quad \mathrm{v}_{1}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{1}\) \((10 \mathrm{a})^{2}=0+2 \mathrm{as}_{1}\) \(\mathrm{s}_{1}=\frac{100 \mathrm{a}^{2}}{2 \mathrm{a}}=50 \mathrm{a}\) Velocity after \(\mathrm{t}_{2}=20\) second \(\mathrm{v}_{2}=\mathrm{u}+\mathrm{at}_{2}\) \(\mathrm{v}_{2}=0+\mathrm{a} \times 20\) \(\mathrm{v}_{2}=20 \mathrm{a}\) And \(\quad \mathrm{v}_{2}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{2}\) \((20 a)^{2}=0+2 a_{2}\) \(s_{2}=s_{1}=\frac{100 a^{2}}{2 a}=50 a\) So, \(\quad \frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{50 \mathrm{a}}{200 \mathrm{a}}\) \(\frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{1}{4}\) \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
141149
An object is moving with a uniform acceleration which is parallel to its instantaneous direction of motion. The displacement (s) - velocity (v) graph of this object is
1
2
3
4
Explanation:
C From Newton's third equation of motion, \(\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as}\) \(\mathrm{v}^{2}=2 \mathrm{as} \quad[\because \mathrm{u}=0]\) We know, Parabola equation, \(x^{2}=2\) ay
AP EAMCET-20.08.2021
Motion in One Dimensions
141150
The relation between time \(t\) and distance \(x\) is \(t\) \(=a x^{2}+b x\), where \(a\) and \(b\) are constants. The acceleration is
1 \(-2 a b v^{2}\)
2 \(2 \mathrm{bv}^{3}\)
3 \(-2 \mathrm{av}^{3}\)
4 \(2 \mathrm{av}^{2}\)
Explanation:
C Let the acceleration be A. \(\Rightarrow \quad \mathrm{t}=\mathrm{ax}^{2}+\mathrm{bx}\) On differentiating the above equation with respect to \(t\). Then, \(\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{ax}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{bx})\) \(1=2 \mathrm{ax} \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{b} \frac{\mathrm{dx}}{\mathrm{dt}}\) \(1=\frac{\mathrm{dx}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Again differentiating the equation number (ii) w.r.t. \(t\) \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{(2 \mathrm{ax}+\mathrm{b}) \frac{\mathrm{d}}{\mathrm{dt}}(1)-1 \cdot \frac{\mathrm{d}}{\mathrm{dt}}(2 \mathrm{ax}+\mathrm{b})}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\because \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{a} \frac{\mathrm{dx}}{\mathrm{dt}}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) \(\therefore \quad \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\therefore \quad \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{-2 \mathrm{av}}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) From equation no. (ii) \(\mathrm{v}=\frac{1}{(2 \mathrm{ax}+\mathrm{b})}\) Acceleration \(=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=-2 \mathrm{av} \times \frac{1}{(2 \mathrm{ax}+\mathrm{b})^{2}}\) Acceleration \(=-2 \mathrm{av} \times \mathrm{v}^{2}\) Acceleration \(=-2 \mathrm{av}^{3}\)
MP PET-2012
Motion in One Dimensions
141151
A car moving with a speed of \(40 \mathrm{~km} / \mathrm{h}\) can be stopped after \(2 \mathrm{~m}\) by applying brakes. If the same car is moving with a speed of \(80 \mathrm{~km} / \mathrm{h}\), what is the minimum stopping distance?
1 \(8 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(4 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
A For first case : \(\mathrm{u}=40 \mathrm{~km} / \mathrm{h}\) \(=\left(40 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) For third equation of motion we have, \(v^{2}=u^{2}+2 a s\) \(0=\left(40 \times \frac{5}{18}\right)^{2}+2 a(2)\) \(0=\left(\frac{200}{18}\right)^{2}+4 a\) \(\mathrm{a}=-\frac{1}{4}\left(\frac{100}{9}\right)^{2}\) For second case :- \(\mathrm{u}=80 \mathrm{~km} / \mathrm{h}\) \(\mathrm{u}=\left(80 \times \frac{5}{18}\right) \mathrm{m} / \mathrm{s}\) \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as}\) \(0=\left(80 \times \frac{5}{18}\right)^{2}+2 \times \mathrm{a} \times \mathrm{s}\) \(2 \mathrm{as}=-\left(\frac{200}{9}\right)^{2}\) Putting the value of ' \(a\) ' in equation (ii) \(-2 \times \frac{1}{4} \times\left(\frac{100}{9}\right)^{2} \times s=-\left(\frac{200}{9}\right)^{2}\) \(\frac{2}{4} \times \frac{100}{9} \times \frac{100}{9} \times s=\left(\frac{200}{9}\right)^{2}\) \(\frac{s}{2}=2 \times 2\) \(s=2 \times 2 \times 2\) Distance, \(\mathrm{s}=8 \mathrm{~m}\)
AIEEE-2003
Motion in One Dimensions
141152
A particle moves along a straight line \(O X\). At a time \(t\) (in seconds) the distance \(x\) (in meters) of the particle from \(O\) is given by \(x=40+12 t-t^{3}\) How long would the particle travel before coming to rest?
1 \(24 \mathrm{~m}\)
2 \(40 \mathrm{~m}\)
3 \(56 \mathrm{~m}\)
4 \(16 \mathrm{~m}\)
Explanation:
C Given that, Distance \(\mathrm{x}=40+12 \mathrm{t}-\mathrm{t}^{3}\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{dx}}{\mathrm{dt}}\) \(\mathrm{v} =\frac{\mathrm{d}}{\mathrm{dt}}\left(40+12 \mathrm{t}-\mathrm{t}^{3}\right)\) \(\mathrm{v} =12-3 \mathrm{t}^{2}\) Final velocity of the particle is zero \((\mathrm{v}=0)\) So, \(\quad 12-3 \mathrm{t}^{2}=0\) \(\mathrm{t}^{2}=\frac{12}{3}\) \(\mathrm{t}=\sqrt{4}\) \(\mathrm{t}=2 \mathrm{sec} .\) Distance covered by the particle before coming to rest is \(\mathrm{x} =40+12 \mathrm{t}-\mathrm{t}^{3}\) \(=40+12 \times 2-2^{3}\) \(=64-8\) \(=56 \mathrm{~m}\)
WE JEE-2012
Motion in One Dimensions
141153
A particle starts its motion from rest under the action of a constant force. If the distance covered in first \(10 \mathrm{~s}\) is \(S_{1}\) and that covered in the first \(20 \mathrm{~s}\) is \(S_{2}\), then
1 \(s_{2}=2 s_{1}\)
2 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
Explanation:
C Particle starts motion from rest, \(\mathrm{u}=0\) Let, acceleration \(=\mathrm{a} \mathrm{m} / \mathrm{s}^{2}\), time \(\mathrm{t}_{1}=10 \mathrm{sec}\) velocity after 10 second \(=\mathrm{v}_{1} \mathrm{~m} / \mathrm{s}\) So, \(\quad \mathrm{v}_{1}=\mathrm{u}+\mathrm{at}_{1}\) \(\mathrm{v}_{1}=0+\mathrm{a} \times 10\) \(\mathrm{v}_{1}=10 \mathrm{a} \mathrm{m} / \mathrm{s}\) And \(\quad \mathrm{v}_{1}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{1}\) \((10 \mathrm{a})^{2}=0+2 \mathrm{as}_{1}\) \(\mathrm{s}_{1}=\frac{100 \mathrm{a}^{2}}{2 \mathrm{a}}=50 \mathrm{a}\) Velocity after \(\mathrm{t}_{2}=20\) second \(\mathrm{v}_{2}=\mathrm{u}+\mathrm{at}_{2}\) \(\mathrm{v}_{2}=0+\mathrm{a} \times 20\) \(\mathrm{v}_{2}=20 \mathrm{a}\) And \(\quad \mathrm{v}_{2}^{2}=\mathrm{u}^{2}+2 \mathrm{as}_{2}\) \((20 a)^{2}=0+2 a_{2}\) \(s_{2}=s_{1}=\frac{100 a^{2}}{2 a}=50 a\) So, \(\quad \frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{50 \mathrm{a}}{200 \mathrm{a}}\) \(\frac{\mathrm{s}_{1}}{\mathrm{~s}_{2}}=\frac{1}{4}\) \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
