A car starts from rest at a stop sign. It accelerates at 2.0 m/ s 2 for 6.9 seco

Question

A car starts from rest at a stop sign. It accelerates at 2.0 m/ s 2 for 6.9 seconds, coasts for 2.5 s , and then slows down at a rate of 1.5 m/ s 2 for the next stop sign.
How far apart are the stop signs? A car starts from rest at a stop sign. It accelerates at 2.0 m/ s 2 for 6.9 seconds, coasts for 2.5 s , and then slows down at a rate of 1.5 m/ s 2 for the next stop sign.
How far apart are the stop signs? A car starts from rest at a stop sign. It accelerates at 2.0 m/ s 2 for 6.9 seconds, coasts for 2.5 s , and then slows down at a rate of 1.5 m/ s 2 for the next stop sign.
How far apart are the stop signs?

Explanation / Answer

Initial velocity of the car Vi = 0 m/s

acceleration a1 = 2.0 m/s2

time t1 = 6.9 sec

It moves with constant speed for time t2 = 2.5 sec

acceleration a2 =0

acceleration a3 = 1.5 m/s2

final velocity Vf =0

So distance covered in first 6.9 sec

d1 = Vi*t + 1/2at2

d1 = 0 + 1/2(2)(6.9)2

d1 = 47.61 m

distance covered while it was moving with constant speed

d2 = V2*t2

where V2 = Vi +at1

V2 = 0 + 2*6.9

V2 = 13.8 m/s

d2 = (13.8)(2.5)

d2 = 34.5 m

now distance covered while it is slowing down

Vf2 = V22 + 2ad3

0 = (13.8)2 +2*(-1.5)d3

d3 = 63.48 m

total distance covered D = d1 +d2 +d3

D = 47.61 + 34.5 +63.48

D = 145.59 m

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