A car starts from rest and accelerates at 5 m/s^2 for 4 s, then maintains that velocity for 3 seconds, then decelerates at the rate of 2m/s^2 for 3 seconds.

Question

a) what is the final speed of the car?
b) how far does the car travel?

a) v = v_0 +at
= 0+5(4)= 20m/s
=0+5(3)= 15m/s
=0+2(3)=6m/s

so I found the veolcity for each of the three parts what would be the final velocity? would it be v-v_0 so 6-20 =-14m/s

b) d=vt
=20*4=80m
=15*3=45m
=6*3=18m

then I add these all up and got 143m as the answer.

Answer 1

No.
The velocity attained in the first period is 20m/s. Then, in the third period, a further deacceleration brings it down 14 m/s.
distance:
The average velocity during the first period is 10m/s, for four seconds: 40m
the average velocity during the second period is 20m/s for 3 seconds, and additional 60 m
The average velocity during the last period is 17 m/s, for three seconds, and additional 17*3 or 51m