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:
\[\begin{align}
  & \text{s = ut + }\frac{1}{2}\text{a}{{\text{t}}^{2}} \\
 & \text{s = 0 + }\frac{1}{2}\text{a}{{\text{t}}^{2}} \\
 & 1500=\frac{1}{2}\times 5\times {{\text{t}}^{2}} \\
 & {{\text{t}}^{2}}=600 \\
 & \text{t = 10}\sqrt{6}\text{ or 24}\text{.49 sec}
\end{align}\]
Now, to calculate the velocity with which distance \[\text{s}\] was covered, we use the equation:
\[\begin{align}
  & \text{v = u + at} \\
 & \text{v = 0 + 5}\times \text{24}\text{.49} \\
 & \text{v = 122}\text{.45 m/sec} \\
\end{align}\]
Finally, the time at which body comes to rest is found using:
\[\begin{align}
  & \text{v = u + at} \\
 & \text{0 = 122}\text{.45}-10\times \text{t} \\
 & \text{t = 12}\text{.245 sec} \\
\end{align}\]
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.