Question

A cheetah can accelerate from $0\,kmh{r^{ - 1}}$ to $90\,kmh{r^{ - 1}}$ in $2.0\,\sec$ while a tiger required $4.5\,\sec$. The ratio of average acceleration of cheetah to tiger is
(A) $\dfrac{9}{2}$
(B) $\dfrac{3}{2}$
(C) $\dfrac{3}{4}$
(D) $\dfrac{9}{4}$

Answer 1

Hint
To determine the ratio of the average acceleration of the cheetah to the tiger is determined by dividing the average of the cheetah by the average acceleration of the tiger. The average acceleration of the cheetah and the tiger is determined by dividing the difference of the final velocity to the initial velocity to the time taken.
The average acceleration of the cheetah is given by,
$\Rightarrow {a_c} = \dfrac{{{V_{{f_c}}} - {V_{{i_c}}}}}{{{t_c}}}$
Where, ${a_c}$ is the average acceleration of the cheetah, ${V_{{f_c}}}$ is the final velocity of the cheetah, ${V_{{i_c}}}$ is the initial velocity of the cheetah and ${t_c}$ is the time taken by the cheetah.
The average acceleration of the tiger is given by,
$\Rightarrow {a_t} = \dfrac{{{V_{{f_t}}} - {V_{{i_t}}}}}{{{t_t}}}$
Where, ${a_t}$ is the average acceleration of the tiger, ${V_{{f_t}}}$ is the final velocity of the tiger, ${V_{{i_t}}}$ is the initial velocity of the tiger and ${t_t}$ is the time taken by the tiger.

Complete step by step answer
Given that, The initial velocity of the cheetah is, ${V_{{i_c}}} = 0\,kmh{r^{ - 1}}$.
The final velocity of the cheetah is, ${V_{{f_c}}} = 90\,kmh{r^{ - 1}}$.
The time taken by the cheetah is, ${t_c} = 2\,\sec$.
The initial velocity of the tiger is, ${V_{{i_t}}} = 0\,kmh{r^{ - 1}}$.
The final velocity of the cheetah is, ${V_{{f_t}}} = 90\,kmh{r^{ - 1}}$.
The time taken by the cheetah is, ${t_t} = 4.5\,\sec$.
Now, The average acceleration of the cheetah is given by,
$\Rightarrow {a_c} = \dfrac{{{V_{{f_c}}} - {V_{{i_c}}}}}{{{t_c}}}\,....................\left( 1 \right)$
By substituting the initial and final velocity of the cheetah and time taken in the equation (1), then
$\Rightarrow {a_c} = \dfrac{{90 - 0}}{2}$
Now the above equation is written as,
$\Rightarrow {a_c} = \dfrac{{90}}{2}$
$\Rightarrow {a_c} = 45$
Now, The average acceleration of the tiger is given by,
$\Rightarrow {a_t} = \dfrac{{{V_{{f_t}}} - {V_{{i_t}}}}}{{{t_t}}}\,........................\left( 2 \right)$
Now substituting the initial and final velocity and time taken by the tiger in the equation (2), then
$\Rightarrow {a_t} = \dfrac{{90 - 0}}{{4.5}}$
Now the above equation is written as,
$\Rightarrow {a_t} = \dfrac{{90}}{{4.5}}$
On dividing the above equation, then
$\Rightarrow {a_t} = 20$
Now, the ratio of the average acceleration of the cheetah to the tiger is,
$\Rightarrow \dfrac{{{a_c}}}{{{a_t}}} = \dfrac{{45}}{{20}}$
On dividing the above equation, then
$\Rightarrow \dfrac{{{a_c}}}{{{a_t}}} = \dfrac{9}{4}$
Hence, the option (D) is the correct answer.

Note
The ratio of the average acceleration of the cheetah to the average acceleration of the tiger has no unit, because both the accelerations are having the same unit, on dividing the unit of both the acceleration gets cancelled. This ratio shows that the cheetah runs $2.25$ times faster than the tiger.