141773
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=\left(t^{3}-6 t^{2}+3 t+4\right) m\) The velocity when the acceleration is zero, is
141774
The position \(x\) of a particle w.r.t. time \(t\) along \(x-\) axis is given by \(x=9 t^{2}-t^{3}\), where \(x\) is in metre and \(t\) in sec. What will be the position of this particle when it achieves maximum speed along the \(+x\) direction?
1 \(32 \mathrm{~m}\)
2 \(54 \mathrm{~m}\)
3 \(81 \mathrm{~m}\)
4 \(24 \mathrm{~m}\)
Explanation:
B Given, \(x=9 t^{2}-t^{3}\) \(\text { speed }(v)=\frac{d x}{d t}=18 t-3 t^{2}\) Maximum velocity at \(\frac{\mathrm{dv}}{\mathrm{dt}}=0\), \(\frac{\mathrm{dv}}{\mathrm{dt}}=18-6 \mathrm{t}\) The time will be- \(0=18-6 \mathrm{t}\) \(\mathrm{t}=3 \mathrm{sec}\) From equation (1) \(x=9(3)^{2}-(3)^{3}\) \(x=81-27\) \(x=54 m\)
AIPMT 2007
Motion in One Dimensions
141776
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=3 t^{3}+7 t^{2}+14 t+5\). The acceleration of the particle at \(t=1 \mathrm{~s}\) is
141777
The position \(x\) of a particle varies with time \(t\), as \(x=a t^{2}-b t^{3}\). The acceleration of the particle will be zero at time \(t\) equals to
1 zero
2 \(\frac{a}{3 b}\)
3 \(\frac{2 \mathrm{a}}{3 \mathrm{~b}}\)
4 \(\frac{a}{b}\)
Explanation:
B Given, \(x=a t^{2}-b t^{3}\) Velocity \((v)=\frac{d x}{d t}=2 a t-3 b t^{2}\) \(a=\frac{d v}{d t}=2 a-6 b t\) According to question \((\mathrm{a}=0)\) \(0=2 a-6 b t\) \(6 b t=2 a\) \(t=\frac{2 a}{6 b} \Rightarrow t=\frac{a}{3 b}\)
AIPMT-1997
Motion in One Dimensions
141778
If the velocity of a particle is \(v=A t+B^{2}\), where \(A\) and \(B\) are constants, then the distance travelled by it between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is
141773
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=\left(t^{3}-6 t^{2}+3 t+4\right) m\) The velocity when the acceleration is zero, is
141774
The position \(x\) of a particle w.r.t. time \(t\) along \(x-\) axis is given by \(x=9 t^{2}-t^{3}\), where \(x\) is in metre and \(t\) in sec. What will be the position of this particle when it achieves maximum speed along the \(+x\) direction?
1 \(32 \mathrm{~m}\)
2 \(54 \mathrm{~m}\)
3 \(81 \mathrm{~m}\)
4 \(24 \mathrm{~m}\)
Explanation:
B Given, \(x=9 t^{2}-t^{3}\) \(\text { speed }(v)=\frac{d x}{d t}=18 t-3 t^{2}\) Maximum velocity at \(\frac{\mathrm{dv}}{\mathrm{dt}}=0\), \(\frac{\mathrm{dv}}{\mathrm{dt}}=18-6 \mathrm{t}\) The time will be- \(0=18-6 \mathrm{t}\) \(\mathrm{t}=3 \mathrm{sec}\) From equation (1) \(x=9(3)^{2}-(3)^{3}\) \(x=81-27\) \(x=54 m\)
AIPMT 2007
Motion in One Dimensions
141776
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=3 t^{3}+7 t^{2}+14 t+5\). The acceleration of the particle at \(t=1 \mathrm{~s}\) is
141777
The position \(x\) of a particle varies with time \(t\), as \(x=a t^{2}-b t^{3}\). The acceleration of the particle will be zero at time \(t\) equals to
1 zero
2 \(\frac{a}{3 b}\)
3 \(\frac{2 \mathrm{a}}{3 \mathrm{~b}}\)
4 \(\frac{a}{b}\)
Explanation:
B Given, \(x=a t^{2}-b t^{3}\) Velocity \((v)=\frac{d x}{d t}=2 a t-3 b t^{2}\) \(a=\frac{d v}{d t}=2 a-6 b t\) According to question \((\mathrm{a}=0)\) \(0=2 a-6 b t\) \(6 b t=2 a\) \(t=\frac{2 a}{6 b} \Rightarrow t=\frac{a}{3 b}\)
AIPMT-1997
Motion in One Dimensions
141778
If the velocity of a particle is \(v=A t+B^{2}\), where \(A\) and \(B\) are constants, then the distance travelled by it between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is
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Motion in One Dimensions
141773
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=\left(t^{3}-6 t^{2}+3 t+4\right) m\) The velocity when the acceleration is zero, is
141774
The position \(x\) of a particle w.r.t. time \(t\) along \(x-\) axis is given by \(x=9 t^{2}-t^{3}\), where \(x\) is in metre and \(t\) in sec. What will be the position of this particle when it achieves maximum speed along the \(+x\) direction?
1 \(32 \mathrm{~m}\)
2 \(54 \mathrm{~m}\)
3 \(81 \mathrm{~m}\)
4 \(24 \mathrm{~m}\)
Explanation:
B Given, \(x=9 t^{2}-t^{3}\) \(\text { speed }(v)=\frac{d x}{d t}=18 t-3 t^{2}\) Maximum velocity at \(\frac{\mathrm{dv}}{\mathrm{dt}}=0\), \(\frac{\mathrm{dv}}{\mathrm{dt}}=18-6 \mathrm{t}\) The time will be- \(0=18-6 \mathrm{t}\) \(\mathrm{t}=3 \mathrm{sec}\) From equation (1) \(x=9(3)^{2}-(3)^{3}\) \(x=81-27\) \(x=54 m\)
AIPMT 2007
Motion in One Dimensions
141776
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=3 t^{3}+7 t^{2}+14 t+5\). The acceleration of the particle at \(t=1 \mathrm{~s}\) is
141777
The position \(x\) of a particle varies with time \(t\), as \(x=a t^{2}-b t^{3}\). The acceleration of the particle will be zero at time \(t\) equals to
1 zero
2 \(\frac{a}{3 b}\)
3 \(\frac{2 \mathrm{a}}{3 \mathrm{~b}}\)
4 \(\frac{a}{b}\)
Explanation:
B Given, \(x=a t^{2}-b t^{3}\) Velocity \((v)=\frac{d x}{d t}=2 a t-3 b t^{2}\) \(a=\frac{d v}{d t}=2 a-6 b t\) According to question \((\mathrm{a}=0)\) \(0=2 a-6 b t\) \(6 b t=2 a\) \(t=\frac{2 a}{6 b} \Rightarrow t=\frac{a}{3 b}\)
AIPMT-1997
Motion in One Dimensions
141778
If the velocity of a particle is \(v=A t+B^{2}\), where \(A\) and \(B\) are constants, then the distance travelled by it between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is
141773
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=\left(t^{3}-6 t^{2}+3 t+4\right) m\) The velocity when the acceleration is zero, is
141774
The position \(x\) of a particle w.r.t. time \(t\) along \(x-\) axis is given by \(x=9 t^{2}-t^{3}\), where \(x\) is in metre and \(t\) in sec. What will be the position of this particle when it achieves maximum speed along the \(+x\) direction?
1 \(32 \mathrm{~m}\)
2 \(54 \mathrm{~m}\)
3 \(81 \mathrm{~m}\)
4 \(24 \mathrm{~m}\)
Explanation:
B Given, \(x=9 t^{2}-t^{3}\) \(\text { speed }(v)=\frac{d x}{d t}=18 t-3 t^{2}\) Maximum velocity at \(\frac{\mathrm{dv}}{\mathrm{dt}}=0\), \(\frac{\mathrm{dv}}{\mathrm{dt}}=18-6 \mathrm{t}\) The time will be- \(0=18-6 \mathrm{t}\) \(\mathrm{t}=3 \mathrm{sec}\) From equation (1) \(x=9(3)^{2}-(3)^{3}\) \(x=81-27\) \(x=54 m\)
AIPMT 2007
Motion in One Dimensions
141776
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=3 t^{3}+7 t^{2}+14 t+5\). The acceleration of the particle at \(t=1 \mathrm{~s}\) is
141777
The position \(x\) of a particle varies with time \(t\), as \(x=a t^{2}-b t^{3}\). The acceleration of the particle will be zero at time \(t\) equals to
1 zero
2 \(\frac{a}{3 b}\)
3 \(\frac{2 \mathrm{a}}{3 \mathrm{~b}}\)
4 \(\frac{a}{b}\)
Explanation:
B Given, \(x=a t^{2}-b t^{3}\) Velocity \((v)=\frac{d x}{d t}=2 a t-3 b t^{2}\) \(a=\frac{d v}{d t}=2 a-6 b t\) According to question \((\mathrm{a}=0)\) \(0=2 a-6 b t\) \(6 b t=2 a\) \(t=\frac{2 a}{6 b} \Rightarrow t=\frac{a}{3 b}\)
AIPMT-1997
Motion in One Dimensions
141778
If the velocity of a particle is \(v=A t+B^{2}\), where \(A\) and \(B\) are constants, then the distance travelled by it between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is
141773
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=\left(t^{3}-6 t^{2}+3 t+4\right) m\) The velocity when the acceleration is zero, is
141774
The position \(x\) of a particle w.r.t. time \(t\) along \(x-\) axis is given by \(x=9 t^{2}-t^{3}\), where \(x\) is in metre and \(t\) in sec. What will be the position of this particle when it achieves maximum speed along the \(+x\) direction?
1 \(32 \mathrm{~m}\)
2 \(54 \mathrm{~m}\)
3 \(81 \mathrm{~m}\)
4 \(24 \mathrm{~m}\)
Explanation:
B Given, \(x=9 t^{2}-t^{3}\) \(\text { speed }(v)=\frac{d x}{d t}=18 t-3 t^{2}\) Maximum velocity at \(\frac{\mathrm{dv}}{\mathrm{dt}}=0\), \(\frac{\mathrm{dv}}{\mathrm{dt}}=18-6 \mathrm{t}\) The time will be- \(0=18-6 \mathrm{t}\) \(\mathrm{t}=3 \mathrm{sec}\) From equation (1) \(x=9(3)^{2}-(3)^{3}\) \(x=81-27\) \(x=54 m\)
AIPMT 2007
Motion in One Dimensions
141776
A particle moves along a straight line such that its displacement at any time \(t\) is given by \(s=3 t^{3}+7 t^{2}+14 t+5\). The acceleration of the particle at \(t=1 \mathrm{~s}\) is
141777
The position \(x\) of a particle varies with time \(t\), as \(x=a t^{2}-b t^{3}\). The acceleration of the particle will be zero at time \(t\) equals to
1 zero
2 \(\frac{a}{3 b}\)
3 \(\frac{2 \mathrm{a}}{3 \mathrm{~b}}\)
4 \(\frac{a}{b}\)
Explanation:
B Given, \(x=a t^{2}-b t^{3}\) Velocity \((v)=\frac{d x}{d t}=2 a t-3 b t^{2}\) \(a=\frac{d v}{d t}=2 a-6 b t\) According to question \((\mathrm{a}=0)\) \(0=2 a-6 b t\) \(6 b t=2 a\) \(t=\frac{2 a}{6 b} \Rightarrow t=\frac{a}{3 b}\)
AIPMT-1997
Motion in One Dimensions
141778
If the velocity of a particle is \(v=A t+B^{2}\), where \(A\) and \(B\) are constants, then the distance travelled by it between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\) is