Question

A scooter can produce a maximum acceleration of $5\text{m/}{{\text{s}}^{2}}$. It’s brakes can produce a maximum retardation of $10\text{m/}{{\text{s}}^{2}}$. Find the maximum time in which it starts from rest, covers a distance of $1.5\text{ km}$ and stops again.

Answer 1

Hint Firstly, using $\text{s = ut + }\frac{1}{2}\text{a}{{\text{t}}^{2}}$, time at which the body starts from rest is calculated. Then, using $\text{v = u + at}$, velocity for covering distance $\text{s}$ is found. Finally, using $\text{v = u + at}$, time in which body stops is calculated.
Formula used: $\text{s = ut + }\frac{1}{2}\text{a}{{\text{t}}^{2}}$ and $\text{v = u + at}$, both are equations of motion.

Complete Step by step solution
Firstly, we will calculate the time at which body starts from rest using equation:

& \text{s = ut + }\frac{1}{2}\text{a}{{\text{t}}^{2}} \\\ & \text{s = 0 + }\frac{1}{2}\text{a}{{\text{t}}^{2}} \\\ & 1500=\frac{1}{2}\times 5\times {{\text{t}}^{2}} \\\ & {{\text{t}}^{2}}=600 \\\ & \text{t = 10}\sqrt{6}\text{ or 24}\text{.49 sec} \end{aligned}$$ Now, to calculate the velocity with which distance $$\text{s}$$ was covered, we use the equation: $$\begin{aligned} & \text{v = u + at} \\\ & \text{v = 0 + 5}\times \text{24}\text{.49} \\\ & \text{v = 122}\text{.45 m/sec} \\\ \end{aligned}$$ Finally, the time at which body comes to rest is found using: $$\begin{aligned} & \text{v = u + at} \\\ & \text{0 = 122}\text{.45}-10\times \text{t} \\\ & \text{t = 12}\text{.245 sec} \\\ \end{aligned}$$ Here, $$\text{a = }-10\text{ m/se}{{\text{c}}^{2}}$$= retardation, $$\text{u = 122}\text{.45 m/sec}$$ and $$\text{v = 0}$$ as body stops. Hence, $$\text{t = 12}\text{.245 sec}$$ is the required answer for this question. **Note** The velocity of the vehicle on a road is never uniform. It’s velocity increases and decreases at random.Since the velocity of the vehicle on the road keeps on changing, it is said to have accelerated motion.