To solve this problem, we can break it down into two parts: the acceleration phase and the deceleration phase.
- Acceleration phase: The car starts from rest and accelerates at a constant rate of 2 m/s^2. We can use the equation of motion:
v = u + at,
where: v = final velocity, u = initial velocity (0 m/s in this case), a = acceleration (2 m/s^2), t = time taken during acceleration phase (unknown).
Since the car travels a distance of 100 m in a total of 20 seconds, we can calculate the time taken during the acceleration phase as follows:
s = ut + (1/2)at^2,
where: s = distance traveled, u = initial velocity, a = acceleration, t = time.
Plugging in the values, we have:
100 = 0 * t + (1/2) * 2 * t^2, 100 = t^2, t = 10 seconds.
Now we can calculate the final velocity at the end of the acceleration phase:
v = u + at, v = 0 + 2 * 10, v = 20 m/s.
- Deceleration phase: During the deceleration phase, the car comes to a rest, so the final velocity is 0 m/s. The initial velocity is 20 m/s (from the end of the acceleration phase). The deceleration is also -2 m/s^2 (negative because it's in the opposite direction of motion).
Using the same equation:
v = u + at,
0 = 20 + (-2) * t, 2t = 20, t = 10 seconds.
So, the deceleration phase also takes 10 seconds.
To find the maximum velocity attained by the car, we take the maximum velocity reached during the acceleration phase, which is 20 m/s.
Therefore, the maximum velocity attained by the car is 20 m/s.