Question

a sports car accelerates from a standing start to 65 mi\h in 4.61s how far can it travel in that time

Answer 1

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 $ft/s^2$

The displacement of the car , $s= ut +\frac{1}{2} at^2$ ,

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 $4.61^2$

                             = 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

#SPJ4