Question

The variation of speed (in m/s) of an object with time (in seconds) is given by the expression $V(t)=V_0-5t+5t^2$

• A. At time t = 0 s, the instantaneous acceleration is zero
• B. At time t = 0 s, there is a deceleration of the object
• C. At time t - 1 s, the object is at rest
• D. At time t = 1 s, the instantaneous acceleration is zero
• E. The distance travelled by the object at time t=1 s is $V_0$ m

Answer 1

Answer: (B)

The velocity of a object is given by equation
$v(t)=v_0-5t+5t^2$
Acceleration
$a=\frac{dv}{dt}=\frac{d}{dt}\left(v_0-5t+5t^2\right)=-5+10t$
at $t=0$ the acceleration of object
$a=-5+10(0)=-5m{s}^{-2}$
So, at time $t=0$, there is deceleration of object.
at $t=1s,v=v_0-5+5=v_{0}m{s}^{-1}$
and $a=-5+10=+5m{s}^{-2}$
So, at $t=0$, there is non zero and positive value of velocity and acceleration.