(A) The velocity $v$ is given as a function of position $s$,where the rate of change of velocity with respect to position is $\frac{dv}{ds} = 5\,m^{-1

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(A) The velocity $v$ is given as a function of position $s$,where the rate of change of velocity with respect to position is $\frac{dv}{ds} = 5\,m^{-1}$.We know that acceleration $a$ is defined as $a = v \frac{dv}{ds}$.Given $v = 20\,m/s$ and $\frac{dv}{ds} = 5\,m^{-1}$.Substituting these values,we get $a = 20 \times 5 = 100\,m/s^2$.

Class 11 Physics Motion in Straight Line Acceleration and its graph

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$A$ particle is moving in a straight line such that its velocity is increasing at $5\,m/s$ per meter. The acceleration of the particle is .......... $m/s^2$ at a point where its velocity is $20\,m/s$.

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$A$ particle is moving in a straight line such that its velocity is increasing at $5\,m/s$ per meter. The acceleration of the particle is .......... $m/s^2$ at a point where its velocity is $20\,m/s$.

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The displacement equation for a particle is $s = 3t^3 + 7t^2 + 14t + 8 \ m$. Its acceleration at time $t = 1 \ s$ is ....... $m/s^2$.

Draw the $x-t$ graphs for positive,negative,and zero acceleration.

If velocity and acceleration are in opposite directions in one-dimensional motion,what happens to the magnitude of velocity?

Acceleration of a particle changes when:

The velocity $(v)-$ time $(t)$ plot of the motion of a body is shown below:The acceleration $(a)-$ time $(t)$ graph that best suits this motion is:

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The velocity $(v)-$ time $(t)$ plot of the motion of a body is shown below:
The acceleration $(a)-$ time $(t)$ graph that best suits this motion is: