When the traffic light turns green, a scooter moves with constant acceleration of 2 ms–2. At same time, a car moving with constant speed of 10 m/s overtakes it. How far beyond the starting point, will the car overtake the scooter?
wouldn't it depend on where the car was when the light turned green?
Assuming...the car was at the greenlight...
distancecar=10t
distancescooter=1/2 *2*t^2
the distances are equal, so
10=t
so you can compute distance with that time.
To find out how far the car will overtake the scooter, we need to determine the time it takes for the car to catch up to the scooter.
Let's say the initial velocity of the scooter is u, and the time taken for the car to overtake the scooter is t. Since the car is moving with a constant speed of 10 m/s, its velocity is v = 10 m/s.
The scooter's velocity can be calculated using the equation:
v_scooter = u + a*t
Given that the scooter moves with constant acceleration of 2 m/s^2, and at the time when the car overtakes the scooter, the scooter has started from rest (u = 0), we have:
v_scooter = 2*t
Now, since the car overtakes the scooter, its distance covered will be equal to the distance covered by the scooter during this time t. We can calculate the distance covered by the scooter using the equation:
s_scooter = u*t + 0.5*a*t^2
With u = 0, and substituting for v_scooter, we have:
s_scooter = 0.5*2*t^2
= t^2
The distance covered by the car will be the same as the distance covered by the scooter:
s_car = s_scooter
= t^2
Now, we need to find t. When the car catches up to the scooter, their velocities will be equal, thus:
v_car = v_scooter
Since v_car = 10 m/s and v_scooter = 2*t, we can set up the equation:
10 = 2*t
Solving for t:
t = 10/2
t = 5 seconds
Now that we have the time taken for the car to overtake the scooter, we can find the distance:
s_car = t^2
= (5)^2
= 25 meters
Therefore, the car will overtake the scooter at a distance of 25 meters beyond the starting point.