Question

A cyclist is going in the positive x-direction at 9m/s. A car initially at rest, accelerates for the first 10 seconds, then it goes with a constant velocity. If the car reaches the cyclist after 15 seconds from the moment the car started moving, find: (a) The acceleration of the car during the first 10 seconds, (b) The velocity of the cyclist with respect to the car when the car reaches the cyclist

Answer 1

1) The car is initially at rest, and it accelerates for the first 10 seconds with acceleration a, so the distance it covers in these first 10 seconds is (in meters)
$d_1 = \frac{1}{2}at^2 = \frac{1}{2}a(10 s)^2 = 50 a$
The velocity the car has reached after these 10 seconds is
$v=at = a (10 s)=10 a$ (3)
Then the car moves for other 5 seconds with this constant velocity (v=10 a) before reaching the cyclist. During this time, the distance it covers is
$d_2 = v t = 10 a \cdot (5 s) =50 a$
So the total distance covered by the car is
$d=d_1 + d_2 = 50 a + 50 a =100 a$ (1)

The cyclist is moving at constant speed of $v=9m/s$, so the distance it covered during the 15 seconds is
$d=vt=(9m/s)(15 s)=135 m$ (2)

And since the car covered the same distance during this time, we can use (1) and (2) to find the acceleration of the car during the first 10 seconds:
$a= \frac{d}{100}= \frac{135}{100} = 1.35 m/s^2$


2) The velocity of the car when it reaches the cyclist is given by (3):
$v_1= 10 a= (10 s)(1.35 m/s^2) = 13.5 m/s$
The velocity of the cyclist is $v_2 = 9m/s$, therefore the velocity of the car relative to the cyclist is
$v' = v_1 - v_2 = 13.5 m/s - 9m/s=4.5 m/s$