Question

A car starts from rest and travels for 5s with a uniform acceleration of 1.5m/s^2. The driver then applies the break, causing a uniform acceleration of -2m/s^2, If the breaks are applied for 3s a) how fast is the car going at the end of the breaking period and b) how far has the car gone?

Answer 1

The initial speed of the car is

The acceleration for the first 5 s is

The displacement of the car in the first five seconds is

And the velocity of the car at 5 s is

Now for the next 3 s, the acceleration of the car is

So, the displacement covered in next 3 s is

So, the total distance traveled by the car from the beginning is

And the velocity of the car after this 3 s is

So, the speed of the car at the end of the breaking period is 1.5 m/s.

And the car has gone a distance of 32.25 m.

Answer 2

To solve this problem, we can break it down into two parts: the initial acceleration phase and the braking phase.

a) Initial acceleration phase: Given: Initial velocity (u) = 0 m/s (car starts from rest) Acceleration (a) = 1.5 m/s^2 Time (t) = 5 s

Using the equation of motion: v = u + at, where v is the final velocity, we can calculate the velocity at the end of the initial acceleration phase:

v = u + at v = 0 + (1.5)(5) v = 7.5 m/s

Therefore, at the end of the initial acceleration phase, the car's velocity is 7.5 m/s.

b) Braking phase: Given: Acceleration (a) = -2 m/s^2 (negative because it opposes the car's motion) Time (t) = 3 s

Using the same equation of motion, we can calculate the distance traveled during the braking phase:

s = ut + (1/2)at^2, where s is the distance traveled

Since the initial velocity for the braking phase is the final velocity from the initial acceleration phase (v = 7.5 m/s), we have:

s = (7.5)(3) + (1/2)(-2)(3)^2 s = 22.5 - 9 s = 13.5 m

Therefore, during the braking phase, the car travels a distance of 13.5 meters.

To summarize: a) At the end of the braking period, the car's speed is 7.5 m/s. b) The car has traveled a distance of 13.5 meters.