Question

The engine of a motorcycle can produce a maximum acceleration of $5m{s^{ - 2}}$. Its brakes can produce a maximum retardation of $10m{s^{ - 2}}$. What is the minimum time in which the motorcycle can cover a distance of $1.5km$ ?

Answer 1

To solve this question, we have to know what is called acceleration. Acceleration Is the rate of change of velocity. Retardation is also the same thing but, retardation is completely at the opposite direction of the acceleration part. so, these two things are vector quantities.

Complete step by step answer:
According to the question, we have acceleration a is equal to $5m/{s^2}$ and maximum retardation is equal to $10m/{s^2}$.We know that the acceleration is equal to the rate of change of velocity. Which is, $a = \dfrac{{v - u}}{t}$
Here we can consider u is equal to zero because the motorcycle was at rest initially. Here we can say, t is the time taken to accelerate.
As we know that, T is the total time than time taken for retardation by an amount of $10m/{s^2}$ be T $-$t.
From the equation of motion we can write retardation is equal to $\dfrac{{v - u}}{t}$.
Here, according to the question, for retardation, final velocity will be zero as motorcycles will come to rest.
Therefore, $- v = - 10(T - t)$
Now, putting the value we get, $- 5t = - 10T = 10t$
So, $T = \dfrac{{3t}}{2}$
The total distance is $1.5km = 1500m$
Therefore,
$\dfrac{1}{2}vT = 1500 \\\ \Rightarrow vT = 3000$
We found,
$v = 5t \\\ \Rightarrow T = \dfrac{{3t}}{2}$
Now,
$5t\dfrac{{3t}}{2} = 3000 \\\ \Rightarrow{t^2} = 400 \\\ \Rightarrow t = 20$
We know that,
$T = \dfrac{{3t}}{2} \\\ \therefore T = 30s$

So, the answer would be $30s$.

Note: We can get confused between the sign of the retardation because this is the same as acceleration. But one thing we have to keep in our mind is that it is exactly at the opposite direction of the acceleration. Acceleration makes an object to achieve more speed and retardation makes an object to get stopped.