A car starts from rest and travels for 5 seconds with a uniform acceleration of 2m/sec^2. The driver then applies the breaks causing a deceleration of 1.5m/sec^2. If the breaks are applied for 4 second, how fast is the car going after the breaking period?
It reaches a speed of 5x2 = 10 m/s after accelerating and then that speed is reduced by 1.5x4 = 6 m/s during deceleration.
10 - 6 = ___ m/s
To solve this problem, we need to break it down into two parts: the initial acceleration and the deceleration.
1. Initial acceleration:
The car starts from rest and undergoes uniform acceleration for 5 seconds with a value of 2 m/sec^2. We can use the following kinematic equation to find the final velocity after this period:
v = u + at
where:
v = final velocity
u = initial velocity (0 m/sec, as the car starts from rest)
a = acceleration (2 m/sec^2)
t = time (5 seconds)
Using the equation, we can calculate the final velocity:
v = 0 + (2 m/sec^2) * (5 sec)
v = 10 m/sec
2. Deceleration:
After the 5-second initial acceleration period, the driver applies the brakes, causing a deceleration of 1.5 m/sec^2. We can use the same kinematic equation to find the final velocity after the deceleration:
v = u + at
where:
v = final velocity (which is the value we need to find)
u = initial velocity (10 m/sec, the final velocity before deceleration)
a = deceleration (-1.5 m/sec^2, negative sign due to deceleration)
t = time (4 seconds)
Using the equation, we can calculate the final velocity:
v = 10 + (-1.5 m/sec^2) * (4 sec)
v = 10 - 6 m/sec
v = 4 m/sec
Therefore, the car is going at a speed of 4 m/sec after the 4-second braking period.