Asked by Babygirl

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.
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!

Answers

Answered by scott
braking time = initial velocity / acceleration

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


speed = acceleration * time
Answered by Babygirl
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?
Answered by Babygirl
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.
Answered by Babygirl
For the second one...
3.2 • 2.4 = 7.68 = 7.7m/s
Would that be correct?
Answered by Babygirl
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
Answered by bobpursley
yes on both.
Answered by Babygirl
Thanks! Can you help with the other question I posted?
Answered by Henry
1. V^2 = Vo^2 + 2a*d = 0,
(22.4)^2 - 7*d = 0,
d = 71.7 m(72m).
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