Question

A speedy rabbit is hopping to the right with a velocity of 4$m{{s}^{-1}}$ when it sees a carrot in the distance. The rabbit speeds up its maximum velocity of 13$m{{s}^{-1}}$ with a constant acceleration of 2$m{{s}^{-2}}$ rightward.
How many seconds does it take the rabbit to speed up from 4$m{{s}^{-1}}$ to 13$m{{s}^{-1}}$ ?

Answer 1

Hint: It is given that the rabbit travels with a constant acceleration. Therefore, use the kinematic equation that gives us the relation between the initial velocity, final velocity, acceleration and time.

Formula used: $v=u+at$

Complete step by step answer:
It is given that the was travelling with a velocity of 4$m{{s}^{-1}}$. When it sees a carrot at a distance, it is said that it speeds up with an acceleration of 2$m{{s}^{-2}}$.
It is given that its maximum velocity is 13$m{{s}^{-1}}$. This means that the velocity of the rabbit when it reaches the carrot is 13$m{{s}^{-1}}$.
We are asked to calculate the time taken for the rabbit to speed up from 4$m{{s}^{-1}}$ to 13$m{{s}^{-1}}$. Let this time be t. To calculate the value of t, we will use the kinematic equation $v=u+at$ …… (i),
where v is the final velocity, u is the initial velocity and a is acceleration.
In this case, u = 4$m{{s}^{-1}}$, v = 13$m{{s}^{-1}}$ and a = 2$m{{s}^{-2}}$.
Substitute the values in (i).
$\Rightarrow 13=4+2t$
$\Rightarrow t=\dfrac{9}{2}=4.5s$.

Therefore, the rabbit speeds up from 4$m{{s}^{-1}}$ to 13$m{{s}^{-1}}$ in 4.5 seconds.

Note: If in the question it was also asked to calculate the displacement of the rabbit when it accelerates from 4$m{{s}^{-1}}$ to 13$m{{s}^{-1}}$, then we can use the kinematic equation $2as={{v}^{2}}-{{u}^{2}}$, where s is the displacement of the rabbit.
Note that all the kinematic equations are applicable only for constant acceleration.