A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.0 seconds, coasts for 2.4 s , and then slows down at a rate of 1.5m/s2 for the next stop sign.

d = 112.8m

the stop signs are 112.8m apart.

A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.0 seconds, coasts for 2.4 s , and then slows down at a rate of 1.5m/s2 for the next stop sign. How far apart are the stop signs?

Explanation:

The motion are in three stages;

1. It accelerates at 2.0 m/s2 for 6.0 seconds

2. It coasts for 2.4 s

3. then slows down at a rate of 1.5m/s2 for the next stop sign.

1. It accelerates at 2.0 m/s2 for 6.0 seconds;

distance d1 = ut + 0.5at^2

u = 0

t = 6.0 s

a = 2.0 m/s^2

d1 = 0 + 0.5×2.0(6^2)

d1 = 36 m

2. It coasts for 2.4 s ;

Final Speed v2 = at = 2 × 6

v2 = 12 m/s

Distance d2 = vt = v2(t2)

T2 = 2.4 s

Substituting the values;

d2 = 12×2.4 = 28.8 m

3. then slows down at a rate of 1.5m/s2 for the next stop sign.

Time taken to decelerate t3 = ∆v/a = 12/1.5 = 8 s

Distance d3 = ut + 0.5at^2

u = 12 m/s

a = -1.5 m/s^2

t = 8 s

d3 = 12(8) +0.5(-1.5 ×8^2)

d3 = 48 m

Total distance d = d1 +d2 +d3

d = 36m + 28.8m + 48m

d = 112.8m

the stop signs are 112.8m apart.

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