A train is moving at 15ms-1. It slows down to 10.5ms over a time of 4s. Calculate the distance it travels before stopping if acc
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The car can travel 220.27 ft in 4.61 s of time.
Given that a sports car accelerates from a standing start to 65 mi/hr.
So the initial velocity = 0 ft/s
The final velocity = 65 mi/hr.
We convert the final velocity from mi/hr. to ft/s.
1 mi/hr = 5280/3600 ft/s
= 1.47 ft/s
Therefore final velocity = 65 x 1.47 = 95.55 ft/s
The time taken by the car = 4.61 s
Now the average acceleration of the car = final velocity - initial velocity / time taken = 95.55 - 0/ 4.61 s = 20.73
The displacement of the car , ,
where u is the initial velocity of the car, t, the time taken by the car and a, the average acceleration.
So, displacement = 0 x 4.61 + 1/2 x 20.73 x
= 220.27 ft
Hence the distance travelled by the sports car in 4.61s is 220.27 ft.
Learn more about the displacement at brainly.com/question/17006213
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