361809
The \(x\) and \(y\) coordinates of the particle at any time are \(x = 5t - 2{t^2}\) and \(y = 10t\) respectively, where \(x\) and \(y\) are in meters and \(t\) in seconds. The acceleration of the particle at \(t = 2s\) is
1 \(5\,m/{s^2}\)
2 \( - 4\,m/{s^2}\)
3 \( - 8\,m/{s^2}\)
4 \(0\)
Explanation:
\({v_x} = 5 - 4t,\,{v_y} = 10\) \({a_x} = - 4,\;{a_y} = 0\) \(\overrightarrow a = {a_x}\hat i + {a_y}\hat j\) \(\overrightarrow a = - 4\hat i\;m/{s^2}\)
NEET - 2017
PHXI04:MOTION IN A PLANE
361810
A particle begins accelerating from rest (at the origin) with a constant acceleration \({a=2 \hat{i}-4 \hat{j}}\) (where \({a}\) has the units \({m / s^{2}}\) ). How far is the particle from the origin at time \({t=1 s}\) ?
1 \({1 m}\)
2 \({2 m}\)
3 \({5 m}\)
4 None of these
Explanation:
\( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} \vec r = \vec ut + \frac{1}{2}\vec a{t^2}\) \( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} \vec r = 0 + \frac{1}{2}(2\hat i - 4\hat j)\left( {{1^2}} \right) = \hat i - 2\hat j\) \( \Rightarrow |\vec r| = \sqrt {{1^2} + {2^2}} = \sqrt 5 \;m\) Displacement from origin is \({\sqrt{5} {~m}}\).
PHXI04:MOTION IN A PLANE
361811
A particle’s velocity changes from \(\left( {2\widehat i + 3\widehat j} \right)m/s\) in to \(\left( {3\widehat i - 2\widehat j} \right)m{\rm{/}}s\) in 2 \(s\). If its mass is 1 \(kg\), the acceleration \(\left( {m{\rm{/}}{s^2}} \right)\) is
1 \( - \left( {\widehat i + 5\widehat j} \right)\)
361812
A bead is free to slide down on a smooth wire rightly stretched between points \(A\) and \(B\) on a vertical circle of radius 10 \(m\) . Find the time taken by the bead to reach point \(B\), if the bead slides from rest from the highest point \(A\) on the circle.
1 \(1\,s\)
2 \(5\,s\)
3 \(2\,s\)
4 \(7\,s\)
Explanation:
\(A B=2 R \cos \theta\)\(\begin{aligned}& A B=\dfrac{1}{2} g \cos \theta t^{2} \quad \Rightarrow \quad 2 R \cos \theta=\dfrac{1}{2} g \cos \theta t^{2} \\& 2 \sqrt{\dfrac{R}{g}}=t \Rightarrow 2 \sqrt{\dfrac{10}{10}}=t=2 {~s}\end{aligned}\)
361809
The \(x\) and \(y\) coordinates of the particle at any time are \(x = 5t - 2{t^2}\) and \(y = 10t\) respectively, where \(x\) and \(y\) are in meters and \(t\) in seconds. The acceleration of the particle at \(t = 2s\) is
1 \(5\,m/{s^2}\)
2 \( - 4\,m/{s^2}\)
3 \( - 8\,m/{s^2}\)
4 \(0\)
Explanation:
\({v_x} = 5 - 4t,\,{v_y} = 10\) \({a_x} = - 4,\;{a_y} = 0\) \(\overrightarrow a = {a_x}\hat i + {a_y}\hat j\) \(\overrightarrow a = - 4\hat i\;m/{s^2}\)
NEET - 2017
PHXI04:MOTION IN A PLANE
361810
A particle begins accelerating from rest (at the origin) with a constant acceleration \({a=2 \hat{i}-4 \hat{j}}\) (where \({a}\) has the units \({m / s^{2}}\) ). How far is the particle from the origin at time \({t=1 s}\) ?
1 \({1 m}\)
2 \({2 m}\)
3 \({5 m}\)
4 None of these
Explanation:
\( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} \vec r = \vec ut + \frac{1}{2}\vec a{t^2}\) \( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} \vec r = 0 + \frac{1}{2}(2\hat i - 4\hat j)\left( {{1^2}} \right) = \hat i - 2\hat j\) \( \Rightarrow |\vec r| = \sqrt {{1^2} + {2^2}} = \sqrt 5 \;m\) Displacement from origin is \({\sqrt{5} {~m}}\).
PHXI04:MOTION IN A PLANE
361811
A particle’s velocity changes from \(\left( {2\widehat i + 3\widehat j} \right)m/s\) in to \(\left( {3\widehat i - 2\widehat j} \right)m{\rm{/}}s\) in 2 \(s\). If its mass is 1 \(kg\), the acceleration \(\left( {m{\rm{/}}{s^2}} \right)\) is
1 \( - \left( {\widehat i + 5\widehat j} \right)\)
361812
A bead is free to slide down on a smooth wire rightly stretched between points \(A\) and \(B\) on a vertical circle of radius 10 \(m\) . Find the time taken by the bead to reach point \(B\), if the bead slides from rest from the highest point \(A\) on the circle.
1 \(1\,s\)
2 \(5\,s\)
3 \(2\,s\)
4 \(7\,s\)
Explanation:
\(A B=2 R \cos \theta\)\(\begin{aligned}& A B=\dfrac{1}{2} g \cos \theta t^{2} \quad \Rightarrow \quad 2 R \cos \theta=\dfrac{1}{2} g \cos \theta t^{2} \\& 2 \sqrt{\dfrac{R}{g}}=t \Rightarrow 2 \sqrt{\dfrac{10}{10}}=t=2 {~s}\end{aligned}\)
361809
The \(x\) and \(y\) coordinates of the particle at any time are \(x = 5t - 2{t^2}\) and \(y = 10t\) respectively, where \(x\) and \(y\) are in meters and \(t\) in seconds. The acceleration of the particle at \(t = 2s\) is
1 \(5\,m/{s^2}\)
2 \( - 4\,m/{s^2}\)
3 \( - 8\,m/{s^2}\)
4 \(0\)
Explanation:
\({v_x} = 5 - 4t,\,{v_y} = 10\) \({a_x} = - 4,\;{a_y} = 0\) \(\overrightarrow a = {a_x}\hat i + {a_y}\hat j\) \(\overrightarrow a = - 4\hat i\;m/{s^2}\)
NEET - 2017
PHXI04:MOTION IN A PLANE
361810
A particle begins accelerating from rest (at the origin) with a constant acceleration \({a=2 \hat{i}-4 \hat{j}}\) (where \({a}\) has the units \({m / s^{2}}\) ). How far is the particle from the origin at time \({t=1 s}\) ?
1 \({1 m}\)
2 \({2 m}\)
3 \({5 m}\)
4 None of these
Explanation:
\( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} \vec r = \vec ut + \frac{1}{2}\vec a{t^2}\) \( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} \vec r = 0 + \frac{1}{2}(2\hat i - 4\hat j)\left( {{1^2}} \right) = \hat i - 2\hat j\) \( \Rightarrow |\vec r| = \sqrt {{1^2} + {2^2}} = \sqrt 5 \;m\) Displacement from origin is \({\sqrt{5} {~m}}\).
PHXI04:MOTION IN A PLANE
361811
A particle’s velocity changes from \(\left( {2\widehat i + 3\widehat j} \right)m/s\) in to \(\left( {3\widehat i - 2\widehat j} \right)m{\rm{/}}s\) in 2 \(s\). If its mass is 1 \(kg\), the acceleration \(\left( {m{\rm{/}}{s^2}} \right)\) is
1 \( - \left( {\widehat i + 5\widehat j} \right)\)
361812
A bead is free to slide down on a smooth wire rightly stretched between points \(A\) and \(B\) on a vertical circle of radius 10 \(m\) . Find the time taken by the bead to reach point \(B\), if the bead slides from rest from the highest point \(A\) on the circle.
1 \(1\,s\)
2 \(5\,s\)
3 \(2\,s\)
4 \(7\,s\)
Explanation:
\(A B=2 R \cos \theta\)\(\begin{aligned}& A B=\dfrac{1}{2} g \cos \theta t^{2} \quad \Rightarrow \quad 2 R \cos \theta=\dfrac{1}{2} g \cos \theta t^{2} \\& 2 \sqrt{\dfrac{R}{g}}=t \Rightarrow 2 \sqrt{\dfrac{10}{10}}=t=2 {~s}\end{aligned}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI04:MOTION IN A PLANE
361809
The \(x\) and \(y\) coordinates of the particle at any time are \(x = 5t - 2{t^2}\) and \(y = 10t\) respectively, where \(x\) and \(y\) are in meters and \(t\) in seconds. The acceleration of the particle at \(t = 2s\) is
1 \(5\,m/{s^2}\)
2 \( - 4\,m/{s^2}\)
3 \( - 8\,m/{s^2}\)
4 \(0\)
Explanation:
\({v_x} = 5 - 4t,\,{v_y} = 10\) \({a_x} = - 4,\;{a_y} = 0\) \(\overrightarrow a = {a_x}\hat i + {a_y}\hat j\) \(\overrightarrow a = - 4\hat i\;m/{s^2}\)
NEET - 2017
PHXI04:MOTION IN A PLANE
361810
A particle begins accelerating from rest (at the origin) with a constant acceleration \({a=2 \hat{i}-4 \hat{j}}\) (where \({a}\) has the units \({m / s^{2}}\) ). How far is the particle from the origin at time \({t=1 s}\) ?
1 \({1 m}\)
2 \({2 m}\)
3 \({5 m}\)
4 None of these
Explanation:
\( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} \vec r = \vec ut + \frac{1}{2}\vec a{t^2}\) \( \Rightarrow \;\;\;{\mkern 1mu} {\kern 1pt} \vec r = 0 + \frac{1}{2}(2\hat i - 4\hat j)\left( {{1^2}} \right) = \hat i - 2\hat j\) \( \Rightarrow |\vec r| = \sqrt {{1^2} + {2^2}} = \sqrt 5 \;m\) Displacement from origin is \({\sqrt{5} {~m}}\).
PHXI04:MOTION IN A PLANE
361811
A particle’s velocity changes from \(\left( {2\widehat i + 3\widehat j} \right)m/s\) in to \(\left( {3\widehat i - 2\widehat j} \right)m{\rm{/}}s\) in 2 \(s\). If its mass is 1 \(kg\), the acceleration \(\left( {m{\rm{/}}{s^2}} \right)\) is
1 \( - \left( {\widehat i + 5\widehat j} \right)\)
361812
A bead is free to slide down on a smooth wire rightly stretched between points \(A\) and \(B\) on a vertical circle of radius 10 \(m\) . Find the time taken by the bead to reach point \(B\), if the bead slides from rest from the highest point \(A\) on the circle.
1 \(1\,s\)
2 \(5\,s\)
3 \(2\,s\)
4 \(7\,s\)
Explanation:
\(A B=2 R \cos \theta\)\(\begin{aligned}& A B=\dfrac{1}{2} g \cos \theta t^{2} \quad \Rightarrow \quad 2 R \cos \theta=\dfrac{1}{2} g \cos \theta t^{2} \\& 2 \sqrt{\dfrac{R}{g}}=t \Rightarrow 2 \sqrt{\dfrac{10}{10}}=t=2 {~s}\end{aligned}\)
