Question

A dirt bike starts up a steep hill with a speed of $5\,\,m{s^{ - 1}}$. Since it is underpowered it slows down at the rate of $0.2\,\,m{s^{ - 2}}$ as it climbs the hill. It cleans the crest of the hill after $15\,\,s$. How far did the dirt bike travel up the hill?
(A) $97.5\,\,m$.
(B) $52.5\,\,m$.
(C) $15\,\,m$.
(D) $75\,\,m$.

Answer 1

The given problem can be solved using the formula derived for the equation of motion, which incorporates the initial velocity, final velocity, time taken and the distance covered by the dirt bike on a steep hill.

Formulae Used:
By using the equation of motion;
${v^2} = {u^2} + 2as$
Where, $v$denotes the final velocity of the dirt bike, $u$ denotes the initial velocity of the dirt bike, $a$ denotes the acceleration of the dirt bike, $s$ denotes the distance covered by the dirt bike

Complete step-by-step solution:
The data given in the problem is:
Initial velocity of the dirt bike, $u = 5\,\,m{s^{ - 1}}$.
Acceleration of the dirt bike, $a = - 0.2\,\,m{s^{ - 2}}$
Time take, $t = 15\,\,s$
By using the equation of motion;
$v = u + at$
By substituting the given values,
Since the dirt bike slows down when it climbs;
$v = 5\,\,m{s^{ - 1}} + \left( { - 0.2\,\,m{s^{ - 2}} \times 15\,\,s} \right) \\\ v = 5 - 3 \\\ v = 2\,\,m{s^{ - 1}} \\\$
By using the equation of motion;
${v^2} = {u^2} + 2as$
Substitute the values of final velocity;
${2^2} = {5^2} + 2 \times \left( { - 0.2} \right) \times s \\\ 4 = 25 - 0.4 \times s \\\$
Since we only need the distance travelled by the dirt bike;
$\- 0.4s = 4 - 25 \\\ s = \dfrac{{21}}{{0.4}} \\\ s = 52.5\,\,m \\\$
Therefore, the distance travelled by the dirt bike is, $s = 52.5\,\,m$
Hence, the option (B) $52.5\,\,m$ is the correct answer.

Note:- We use the value of the acceleration as negative because when the dirt bike climbs the steep hill, it is underpowered and thus the speed of the bike decreases instead of increasing. Since the speed decreases the acceleration of the bike also decreases.