What's the distance between the stoplights?

Question

A car starts from rest at a stoplight and in 8 seconds reaches a speed of 72 km/h. It goes on at that speed for 20 seconds to then stops at another stoplight in 4 seconds. What's the distance between the stoplights?

Answer 1

72 km/h = 72000 m/hr = 72000 m / 3600 s = 20m/s

So if we assume uniform acceleration first 8 seconds, a = 20/8 = 2.5 m/s/s

s = ut + 0.5at^2, u=0, a = 2.5, t=8 giving s = 80m in the first 8 seconds

Next 20 seconds, speed is constant so distance = 20m/s x 20s = 400m

Last 4 seconds, v^2 = u^2 + 2as

v=0 as car stops. u = 20, a = - 20/4 = -5 as car stops in 4s

so 0 = 20^2 -5x2xs giving s = 40m

So distance required = 80 + 400 + 40 = 520m

Answer 2

If your teacher has introduced velocity-time graphs and asked you to relate them to the motion of objects, using a graph of speed vs time to analyze this problem is worthwhile. Remember that slope on a velocity-time graph corresponds to acceleration and that area on a velocity-time corresponds to displacement. So we should try to depict this motion with a velocity graph and then calculate the area.

The speed starts at zero and steadily builds to 20 m/s. (20 m/s = 72km / hr) so lets start our graph at t = 0, v = 0 m/s and have a straight line with positive slope for 8 sec.

Then the speed is constant, so from t = 8 sec to t = 28 sec, draw a horizontal line connected to the speed at t = 8 sec.

Then the speed decreases to zero over 4 seconds, so finish the graph by drawing a downward sloping line from 20 m/s to 0 m/s from t = 28 sec to t = 32 sec. The whole graph should look like a trapezoid with base 32 sec, height 20 m/s and a top base that is 20 sec long. The area of this trapezoid represents the distance travelled. You can calculate this area as either using the formula average base times height or as a triangle plus a rectangle plus another triangle.

The first way gives:

(32 s + 20 s)/2 * 20 m/s

26 s * 20 m/s

520 m

The second way would have terms:

0.5(4 s)(20m/s) + (20 s)(20 m/s) + 0.5(8 s)(20 m/s)

40 m + 400 m + 80 m

520 m

Answer 3

72km/h = 20m/s

Total distance = 20/2·8 + 20·20 + 20/2·4 = 520 m