Question

the velocity V of a body moving along a straight line varies with time varies with time t as v = 2t ^ 2 into e ^ -t where v is in metres per second and t is in second the acceleration of body is zero at t=

Answer 1

the velocity v of a body moving along a straight line varies with time t as $v=2t^2e^{-t}$

we know, acceleration is the rate of change of velocity per unit time,
e.g., $a=\frac{dv}{dt}$

so, differentiate v with respect to t,
$\frac{dv}{dt}=\frac{d\left(2t^2.e^{-t}\right)}{dt}\\\\\\=2t^2\frac{d(e^{-t})}{dt}+e^{-t}\frac{d2t^2}{dt}\\\\\\=-2t^2.e^{-t}+4te^{-t}=2t.e^{-t}(t-2)$

hence, $a=2t.e^{-t}(t-2)$
acceleration will be zero ,
$a=2t.e^{-t}(t-2)=0$
t = 2 , 0

hence, acceleration of the body is zero at t = 0 and 2.

Answer 2

velocity (v) = dv/DT
acceleration (a) = dv/DT
a=4te^-t-2e^-t t^2
at t=0,a =0
at t=0 , a=0