A subway train is traveling at a rate of 22.4 m/s. Brakes are applied and it slows down at a constant rate of 3.5 m/s^2 until it stops at a station. Find the total distance traveled while braking.

Question

72 m
142 m
274 m
163 m

A cyclist is stopped at a traffic light. When the light turns green, the cyclist accelerates at 3.2 m/s^2. After 2.4 seconds, what is the cyclist’s speed?
15 m/s
7.7 m/s
5.6 m/s
0.75 m/s

Please help guys!

Answer 1

braking time = initial velocity / acceleration

distance traveled = (initial velocity / 2) * braking time

speed = acceleration * time

Answer 2

So the first one would be...

22.4/3.5 = 6.4
22.4/2 = 11.2 • 6.4 = 71.68 = 72
So 72 should be the answer right?

Answer 3

By the way, thanks so much for providing the formulas! My lessons don’t even give me the formulas, so I never know what to do.

Answer 4

For the second one...
3.2 • 2.4 = 7.68 = 7.7m/s
Would that be correct?

Answer 5

Could you help with this as well?

The “reaction time” of the average automobile driver is about 0.7 s. (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of 12 ft/s^2 compute the total distance(in feet) covered in coming to a stop after a signal is observed from an initial velocity of 55 mi/h. Let 1 mi/h= 1.466 ft/s.
271 ft
327 ft
113 ft
414 ft

Answer 6

yes on both.

Answer 7

Thanks! Can you help with the other question I posted?

Answer 8

1. V^2 = Vo^2 + 2a*d = 0,
(22.4)^2 - 7*d = 0,
d = 71.7 m(72m).