Question

A motorcyclist situated at the origin is located at a distance of 12 m behind a car. At t = 0, the motorcyclist starts moving with a constant velocity v = 8 m/s and same time the car starts accelerating from rest with a = $m{s}^{-2}$. (a) When and where do they meet? (b) Draw the position-time and velocity-time relations of car and motorcycle on the same graph.

Answer 1

Let the car and motorcycle meet, after a time t. During the time, the motorcycle covers a distance dm (say). Position of motorcycle, $x_m=d_m$ = vt ...(i)
During the time t, the car moves through a distance
$d_c=1/2a{t}^2$ ...(ii)
When both motorcycle and car meet their positions will be the same. From (i) and (ii), $x_m=x_c$.
We have vt = 12 + (1/2)$a{t}^2$
Then substituting v = 8 m/s and a = 2 $m{s}^{-2}$ , we have $t^2$ - 8t + 12 = 0
This yields two real values of time,
i.e t = 2 s and 6 s.
The following conclusion can be made from the above mathematical proceedings.
At t=2 s the motorcycle approaches the car.
The position of motorcyclist $(x_m{)}_{t=2s}=8\times 2$ = 16 m
At t = 6 s, the car overtakes the motorcycle.
The position of the motorcyclist $(x_m{)}_{t=6s}=8\times 6$ = 48 m.
At t = 2 s, the motorcycle overtakes the car. Then it diverges (moves away from the car) till the car acquires a velocity equal to that of the motorcycle at f = 4 s. Then the distance of separation will gradually decreases as the car moves faster than the motorcycle after 4 s. In consequence, at t=6 s, the car overtakes the motorcycle. After that, the car will take a lead leaving the motorcycle behind it.