The velocity v of a body moving along a strai | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The velocity is given by $v = 2 t^2 e^{-t}$.Acceleration $a$ is the rate of change of velocity with respect to time,$a = \frac{dv}{dt}$.Using the

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

Vedclass · Answer

(D) The velocity is given by $v = 2 t^2 e^{-t}$.Acceleration $a$ is the rate of change of velocity with respect to time,$a = \frac{dv}{dt}$.Using the product rule for differentiation: $a = 2 \left[ t^2 \frac{d}{dt}(e^{-t}) + e^{-t} \frac{d}{dt}(t^2) \right]$.$a = 2 [ t^2 (-e^{-t}) + e^{-t} (2t) ]$.$a = 2 e^{-t} (2t - t^2)$.For acceleration to be zero,set $a = 0$:$2 e^{-t} (2t - t^2) = 0$.Since $2 e^{-t}$ is never zero for any finite $t$,we must have $2t - t^2 = 0$.$t(2 - t) = 0$.This gives $t = 0 \ s$ or $t = 2 \ s$.Thus,the acceleration is zero at $t = 0 \ s$ and $t = 2 \ s$.

The velocity $v$ of a body moving along a straight line varies with time $t$ as $v = 2 t^2 e^{-t}$,where $v$ is in $m/s$ and $t$ is in $s$. The acceleration of the body is zero at $t =$

Similar Questions

The velocity-time graph for a particle moving along the $x$-axis is shown in the figure. The corresponding displacement-time graph is correctly shown by

$A$ lift is going up. The total mass of the lift and the passenger is $1500\, kg$. The variation in the speed of the lift is as given in the graph. The height to which the lift takes the passenger is ............ $m$.

Which of the following options is correct for the object having a straight-line motion represented by the following graph?

The $x-t$ graph of a particle moving along a straight line is shown in the figure. The speed-time graph of the particle is correctly shown by:

The diagram shows the variation of $1/v$ (where $v$ is the velocity of the particle) with respect to time $t$. Using the details given in the graph,the instantaneous acceleration at time $t = 3\,s$ will be equal to $...........\,m/s^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module