Question

# A car starts from rest and travels for 7.5 s with a uniform acceleration of +2.8...

A car starts from rest and travels for 7.5 s with a uniform acceleration of +2.8 m/s2. The driver then applies the brakes, causing a uniform acceleration of −2.5 m/s2.If the brakes are applied for 1.0 s, determine each of the following.

(a) How fast is the car going at the end of the braking period? m/s

(b) How far has the car gone? m

First lets compute the distance from rest and before applying the brakes and the velocity at the instant of applying the brakes:
For distance:
d=Vt+1/2at^2
where V=starting velocity
d=(0)(7.5)+1/2(2.8)(7.5)^2
d=78.75 meter
For the velocity at the instant of applying the brakes:
v=V+at
v=0+(2.8)(7.5)
v=21 m/s

Then the driver applies the brakes for deceleration of 2.5 m/s^2 for 1 seconds

the final velocity will be

V(final)=v+AT , where A= deceleration of 2.5 m/s^2 and T= time of braking of 1 s
V(final)=21+(-2.5)(1) ---> A is negative because it is deceleration or acceleration at negative direction
therefore
V(final)=18.5 m/s

The distance from the instant of braking to the final position will be:

D=vT+1/2AT^2
D=(21)(1)+1/2(-2.5)(1)^2
D=19.75 m

Therefore the total distance is
total distance=d+D
total distance=78.75+19.75
total distance=98.5 meters

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