Answer:
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?
Step-by-step 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.