Question

# A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 7.0...

A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 7.0 seconds, coasts For 2.7 s, and then slows down at a rate of 1.5 m/s2 for the next stop sign. How far apart are the stop signs?

Solution: Distance travelled during acceleration of 2.0 m/s2 :

x1 = ut +0.5at^2 [Using Newton's law]

Where, u is initial velocity = 0 m/s and t = 7 second

=> x1 = 0.5*2*7^2

=> x1 = 49 m.

Final velocity (v) = u + at

=> v = 0 + 2*7

=> v = 14 m/s

Now, Distance travelled at constant speed of v = 14 m/s and time (t) = 2.7 s is calculated as:

x2 = 14*2.7

=> x2 = 37.8 m

And, distance travelled during slow down of acceleration (a) = 1.5 m/s^2 and with inital velocity u = 14 m/s and final velocity (v) = 0 (as it will come to rest ) is calculated as:

v^2 = u^2 + 2ax3

=> 0 = 14^2 + 2*(-1.5)(x3)

=> x3 = 65.33 m.

Thus, total distance travelled = x1 + x2+ x3 = 49 + 37.8 + 65.33 = 152.13 m. Answer

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