362158
On turning a corner, a motorist rushing at \(40\;m{\rm{/}}s\), finds a child on the road \(108\;m\) ahead. He instantly stops the engine and applies the brakes so as to stop it within \(1\;m\) of the child, what time is required to stop it?
362159
A particle moves along the \(X\) - axis. The position \(x\) of a particle w.r.t time from origin given by \(x=b_{0}+b_{1} t+b_{2} t^{2}\). The acceleration of particle is
1 \(b_{0}\)
2 \(b_{1}\)
3 \(b_{2}\)
4 \(2 b_{2}\)
Explanation:
Position of particle, \(x=b_{0}+b_{1} t+b_{2} t^{2}\) Velocity, \(v=\left(\dfrac{d x}{d t}\right)=b_{1} 2 b_{2} t\) Acceleration, \(|a|=\dfrac{d^{2} x}{d t^{2}}=2 b_{2}\)
AIIMS - 2012
PHXI03:MOTION IN A STRAIGHT LINE
362160
The motion of a particle along a straight line is described by equation \(x = 8 + 12t - {t^3},\) where, \(x\) is in metre and \(t\) in sec. The retardation of the particle when its velocity becomes zero, is
1 \(12\,m{s^{ - 2}}\)
2 \({\rm{Zero}}\)
3 \(6\,m{s^{ - 2}}\)
4 \(24\,m{s^{ - 2}}\)
Explanation:
Given, \(x = 8 + 12t - {t^3}\) we know \(v = \frac{{dx}}{{dt}}\) and acceleration \(a = \frac{{dv}}{{dt}}\) So,\(v = 12 - 3{t^2}\) and \(a = - \,6t\) At \(t = 2s\) \(v = 0\) and \(a = - \,6 \times 2\) \(a = - \,12\,m/{s^2}\) So, retardation of the particle \( = \,12\,m/{s^2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362161
A particle moves along a straight line such that its displacement at any point time \(t\) is given by \(s = 3{t^3} + 7{t^2} + 14t + 5\). The acceleration of the particle at \(t = 1\,s\) is
362158
On turning a corner, a motorist rushing at \(40\;m{\rm{/}}s\), finds a child on the road \(108\;m\) ahead. He instantly stops the engine and applies the brakes so as to stop it within \(1\;m\) of the child, what time is required to stop it?
362159
A particle moves along the \(X\) - axis. The position \(x\) of a particle w.r.t time from origin given by \(x=b_{0}+b_{1} t+b_{2} t^{2}\). The acceleration of particle is
1 \(b_{0}\)
2 \(b_{1}\)
3 \(b_{2}\)
4 \(2 b_{2}\)
Explanation:
Position of particle, \(x=b_{0}+b_{1} t+b_{2} t^{2}\) Velocity, \(v=\left(\dfrac{d x}{d t}\right)=b_{1} 2 b_{2} t\) Acceleration, \(|a|=\dfrac{d^{2} x}{d t^{2}}=2 b_{2}\)
AIIMS - 2012
PHXI03:MOTION IN A STRAIGHT LINE
362160
The motion of a particle along a straight line is described by equation \(x = 8 + 12t - {t^3},\) where, \(x\) is in metre and \(t\) in sec. The retardation of the particle when its velocity becomes zero, is
1 \(12\,m{s^{ - 2}}\)
2 \({\rm{Zero}}\)
3 \(6\,m{s^{ - 2}}\)
4 \(24\,m{s^{ - 2}}\)
Explanation:
Given, \(x = 8 + 12t - {t^3}\) we know \(v = \frac{{dx}}{{dt}}\) and acceleration \(a = \frac{{dv}}{{dt}}\) So,\(v = 12 - 3{t^2}\) and \(a = - \,6t\) At \(t = 2s\) \(v = 0\) and \(a = - \,6 \times 2\) \(a = - \,12\,m/{s^2}\) So, retardation of the particle \( = \,12\,m/{s^2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362161
A particle moves along a straight line such that its displacement at any point time \(t\) is given by \(s = 3{t^3} + 7{t^2} + 14t + 5\). The acceleration of the particle at \(t = 1\,s\) is
362158
On turning a corner, a motorist rushing at \(40\;m{\rm{/}}s\), finds a child on the road \(108\;m\) ahead. He instantly stops the engine and applies the brakes so as to stop it within \(1\;m\) of the child, what time is required to stop it?
362159
A particle moves along the \(X\) - axis. The position \(x\) of a particle w.r.t time from origin given by \(x=b_{0}+b_{1} t+b_{2} t^{2}\). The acceleration of particle is
1 \(b_{0}\)
2 \(b_{1}\)
3 \(b_{2}\)
4 \(2 b_{2}\)
Explanation:
Position of particle, \(x=b_{0}+b_{1} t+b_{2} t^{2}\) Velocity, \(v=\left(\dfrac{d x}{d t}\right)=b_{1} 2 b_{2} t\) Acceleration, \(|a|=\dfrac{d^{2} x}{d t^{2}}=2 b_{2}\)
AIIMS - 2012
PHXI03:MOTION IN A STRAIGHT LINE
362160
The motion of a particle along a straight line is described by equation \(x = 8 + 12t - {t^3},\) where, \(x\) is in metre and \(t\) in sec. The retardation of the particle when its velocity becomes zero, is
1 \(12\,m{s^{ - 2}}\)
2 \({\rm{Zero}}\)
3 \(6\,m{s^{ - 2}}\)
4 \(24\,m{s^{ - 2}}\)
Explanation:
Given, \(x = 8 + 12t - {t^3}\) we know \(v = \frac{{dx}}{{dt}}\) and acceleration \(a = \frac{{dv}}{{dt}}\) So,\(v = 12 - 3{t^2}\) and \(a = - \,6t\) At \(t = 2s\) \(v = 0\) and \(a = - \,6 \times 2\) \(a = - \,12\,m/{s^2}\) So, retardation of the particle \( = \,12\,m/{s^2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362161
A particle moves along a straight line such that its displacement at any point time \(t\) is given by \(s = 3{t^3} + 7{t^2} + 14t + 5\). The acceleration of the particle at \(t = 1\,s\) is
362158
On turning a corner, a motorist rushing at \(40\;m{\rm{/}}s\), finds a child on the road \(108\;m\) ahead. He instantly stops the engine and applies the brakes so as to stop it within \(1\;m\) of the child, what time is required to stop it?
362159
A particle moves along the \(X\) - axis. The position \(x\) of a particle w.r.t time from origin given by \(x=b_{0}+b_{1} t+b_{2} t^{2}\). The acceleration of particle is
1 \(b_{0}\)
2 \(b_{1}\)
3 \(b_{2}\)
4 \(2 b_{2}\)
Explanation:
Position of particle, \(x=b_{0}+b_{1} t+b_{2} t^{2}\) Velocity, \(v=\left(\dfrac{d x}{d t}\right)=b_{1} 2 b_{2} t\) Acceleration, \(|a|=\dfrac{d^{2} x}{d t^{2}}=2 b_{2}\)
AIIMS - 2012
PHXI03:MOTION IN A STRAIGHT LINE
362160
The motion of a particle along a straight line is described by equation \(x = 8 + 12t - {t^3},\) where, \(x\) is in metre and \(t\) in sec. The retardation of the particle when its velocity becomes zero, is
1 \(12\,m{s^{ - 2}}\)
2 \({\rm{Zero}}\)
3 \(6\,m{s^{ - 2}}\)
4 \(24\,m{s^{ - 2}}\)
Explanation:
Given, \(x = 8 + 12t - {t^3}\) we know \(v = \frac{{dx}}{{dt}}\) and acceleration \(a = \frac{{dv}}{{dt}}\) So,\(v = 12 - 3{t^2}\) and \(a = - \,6t\) At \(t = 2s\) \(v = 0\) and \(a = - \,6 \times 2\) \(a = - \,12\,m/{s^2}\) So, retardation of the particle \( = \,12\,m/{s^2}\)
PHXI03:MOTION IN A STRAIGHT LINE
362161
A particle moves along a straight line such that its displacement at any point time \(t\) is given by \(s = 3{t^3} + 7{t^2} + 14t + 5\). The acceleration of the particle at \(t = 1\,s\) is
