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!
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?
For the second one...
3.2 • 2.4 = 7.68 = 7.7m/s
Would that be correct?
braking time = initial velocity / acceleration
distance traveled = (initial velocity / 2) * braking time
speed = acceleration * time
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.
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
yes on both.
Thanks! Can you help with the other question I posted?
1. V^2 = Vo^2 + 2a*d = 0,
(22.4)^2 - 7*d = 0,
d = 71.7 m(72m).
To find the total distance traveled while braking, we need to find the time it takes for the subway train to come to a stop. We can use the formula:
v = u + at
where v is the final velocity (0 m/s as the train comes to a stop), u is the initial velocity (22.4 m/s), a is the acceleration (-3.5 m/s^2), and t is the time taken.
Rearranging the formula, we get:
t = (v - u) / a
Plugging in the values, we have:
t = (0 - 22.4) / -3.5
t = 6.4 seconds
Now, to calculate the distance traveled while braking, we can use the formula:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity (22.4 m/s), a is the acceleration (-3.5 m/s^2), and t is the time taken (6.4 seconds).
Plugging in the values, we have:
s = (22.4 * 6.4) + (1/2)(-3.5)(6.4^2)
s = 142.4 - ((1/2)(-3.5)(40.96))
s = 142.4 - (-56)
s = 142.4 + 56
s = 198.4 meters
Therefore, the total distance traveled while braking is 198.4 meters.
For the second question, to find the cyclist's speed after 2.4 seconds of acceleration, we can use the formula:
v = u + at
where v is the final velocity, u is the initial velocity (0 m/s as the cyclist is stopped), a is the acceleration (3.2 m/s^2), and t is the time taken (2.4 seconds).
Plugging in the values, we have:
v = 0 + (3.2 * 2.4)
v = 7.68 m/s
Therefore, the cyclist's speed after 2.4 seconds of acceleration is 7.68 m/s.