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a person is driving unsafely on the highway at a speed of 145km/h and has to slam on the brakes in order to avoid a collision....Question
Alex is driving unsafely on the highway at a speed of 123km/h and has to slam on the brakes in order to avoid a collision. The brakes can slow down the car at a rate of 13.6m/s^2.
a) from the moment the car starts to slow down, how long will it take to stop?
b) how far will the car travel while slowing down?
a) from the moment the car starts to slow down, how long will it take to stop?
b) how far will the car travel while slowing down?
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Answered by
GPT 3.5
To solve this problem, we can use the equations of motion.
a) From the moment the car starts to slow down, we want to find the time it takes to stop.
Using the formula v = u + at, where:
v = final velocity (0 m/s, as the car stops)
u = initial velocity (123 km/h, which is equivalent to 123,000 m/3600 s = 34.17 m/s)
a = acceleration (-13.6 m/s^2, as it is slowing down)
t = time
Rearranging the equation to solve for time, we have:
0 = 34.17 + (-13.6)t
-34.17 = -13.6t
t ≈ 2.51 seconds
Therefore, it will take approximately 2.51 seconds for the car to stop.
b) To find the distance the car will travel while slowing down, we can use the equation s = ut + (1/2)at^2, where:
s = distance
u = initial velocity (34.17 m/s)
t = time taken to stop (2.51 s)
a = acceleration (-13.6 m/s^2)
Substituting the values into the equation:
s = 34.17(2.51) + (1/2)(-13.6)(2.51)^2
s ≈ 42.93 meters
Therefore, the car will travel approximately 42.93 meters while slowing down.
a) From the moment the car starts to slow down, we want to find the time it takes to stop.
Using the formula v = u + at, where:
v = final velocity (0 m/s, as the car stops)
u = initial velocity (123 km/h, which is equivalent to 123,000 m/3600 s = 34.17 m/s)
a = acceleration (-13.6 m/s^2, as it is slowing down)
t = time
Rearranging the equation to solve for time, we have:
0 = 34.17 + (-13.6)t
-34.17 = -13.6t
t ≈ 2.51 seconds
Therefore, it will take approximately 2.51 seconds for the car to stop.
b) To find the distance the car will travel while slowing down, we can use the equation s = ut + (1/2)at^2, where:
s = distance
u = initial velocity (34.17 m/s)
t = time taken to stop (2.51 s)
a = acceleration (-13.6 m/s^2)
Substituting the values into the equation:
s = 34.17(2.51) + (1/2)(-13.6)(2.51)^2
s ≈ 42.93 meters
Therefore, the car will travel approximately 42.93 meters while slowing down.
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