A car starts from rest and travels for 5.0 seconds with uniform acceleration of 1.5 m/s^2. The driver then…?

Question

A car starts from rest and travels for 5.0 seconds with a uniform acceleration of 1.5 m/s². The driver then applies the brakes, resulting in a uniform acceleration of -2.0 m/s². If the brakes are applied for 3.0 seconds, what is the car's speed at the end of the braking period, and how far has it traveled during this time? I worked through this problem for homework and would like to verify my calculations and answer.

Anonymous · Accepted Answer

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)(5)+1/2(1.5)(5)^2

d=18.75 meter

For the velocity at the instant of applying the brakes:

v=V+at

v=0+(1.5)(5)

v=7.5 m/s

Then the driver applies the brakes for deceleration of 2.0 m/s^2 for 3 seconds

the final velocity will be

V(final)=v+AT where A= deceleration of 2.0 m/s^2 and T= time of braking of 3 s

V(final)=7.5+(-2)(3) ---> A is negative because it is deceleration or acceleration at negative direction

therefore

V(final)=1.5 m/s

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

D=vT+1/2AT^2

D=(7.5)(3)+1/2(-2)(3)^2

D=22.5+1/2(-2)(9)

D=22.5-9

D=13.5 m

Therefore the total distance is

total distance=d+D

total distance=18.75+13.5

total distance=32.25 meters

Anonymous · Answer

effortless in simple terms use the acceleration equation (vf = vi +at) for the first 7 seconds then use it for the braking subtract both solutions and also you get section a. for b in simple terms use the acceleration equation that includes a distance ingredient to it. (d = vit + a million/2at^2)

Anonymous · Answer

U=0t=5seca=1.5m/s^2V=U+at=0+1.5*5=7.5m/sDitance S=Ut +1/2*at^2             =0+1/2 *1.5*5^2              18.75mNow retardationu=7.5m/s   (last vel in acceleration)a=-2m/s^2t=3secv=u-at7.5-2*31.5m/s (Last velocity)s=ut-1/2at^2=7.5*3-1/2*2*3^2=  22.5-9=13.5m(Last velocity)= 1.5m/sTotal distance=18.75+13.5                        =32.25m

Zella Bode · Answer

10.5 m/s for final velocity
78 m distance

Markus O'Hara

A car starts from rest and travels for 5.0 seconds with uniform acceleration of 1.5 m/s^2. The driver then…?

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