A car starts from rest and travels for 6.2 s
with a uniform acceleration of +2.4 m/s2.
The driver then applies the brakes, causing a
uniform deceleration of 1.2 m/s2.
If the brakes are applied for 2.5 s, how fast
is the car going at the end of the braking
period?
To find the final speed of the car at the end of the braking period, we can break down the problem into two stages: the acceleration stage and the deceleration stage.
Stage 1: Acceleration
Given:
Initial velocity (u) = 0 m/s (car starts from rest)
Acceleration (a) = +2.4 m/s^2
Time (t) = 6.2 s
We can use the equation of motion: v = u + at, where v is the final velocity after time t.
Plugging in the given values, we get:
v1 = 0 + 2.4 * 6.2
v1 = 14.88 m/s
So, the car's speed at the end of the acceleration stage is 14.88 m/s.
Stage 2: Deceleration
Given:
Deceleration (a) = -1.2 m/s^2 (negative because it opposes the car's motion)
Time (t) = 2.5 s
Using the same equation of motion, v = u + at, we can calculate the final speed (v2) at the end of the deceleration stage.
Plugging in the values, we get:
v2 = v1 + (-1.2) * 2.5
v2 = 14.88 - 3
v2 = 11.88 m/s
Therefore, at the end of the braking period, the car is going at a speed of 11.88 m/s.