A bus accelerates at 1.5 m/s 2 from rest for 12 s. It then travels at constant s

Question

A bus accelerates at 1.5 m/s2 from rest for 12 s. It then travels at constant speed for 26 s, after which it slows to a stop with an acceleration of magnitude 1.6 m/s2.

1)

(a) What is the total distance that the bus travels?
km

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2)

(b) What was its average velocity?
m/s

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Explanation / Answer

distance travelled in first part
S1 = ut+0.5at^2 = 0 + 0.5*1.5*12^2 = 108 m
distaNCE TRAVELLED IN SECOND PART
speed v = u+at = 0+1.5*12 = 18 m/s
s2 = 18*26 = 468 m
distance travelled in third part
S3 = v^2-u^2/2a = - 18^2/(-2*1.6) = 101.25 m
total distance covered =S1+s2+s3 = 108+468+101.25 = 677.25 m
2)
time taken to stop in third part
0 = 18 -1.6 t = 11.25 sec
t =
average velocity = 677.25/(12+26+11.25) = 13.75 m/s

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