A high-performance sports car can go from 0 to 100 mph in 5.8 s. (Assume the car travels in the positive direction. Indicate the direction with the sign of your answer.) (a) What is the car's average acceleration (in SI units; meter-kilogram-second)The same car can come to a complete stop from 34 m/s in 5.1 s. What is its average acceleration?

a) 7.71 m/s², b) - 6.67 m/s²

Explanation: first thing to note is that

1 mile = 1609.34

1 hour = 3600s

Hence, 100mph to m/s = (100 × 1609.34)/3600 = 44.71 m/s

Initial velocity (u) = 0, final velocity (v) = 44.71 m/s, t = 5.8s, a = acceleration =?

By using newton's laws of motion

v = u + at

44.71 = 0 + a(5.8)

44.71 = 5.8a

a = 44.71/5.8

a = 7.71 m/s²

Question b)

The car is completing a stop which implies that the car is coming to rest, and when a car is coming to rest, the final velocity (v) is zero.

Hence u = 34 m/s, v = 0, a =?, t = 5.1 s

v = u + at

0 = 34 + a(5.1)

a(5.1) = - 34

a = - 34/5.1

a = - 6.67 m/s².

The negative sign beside the acceleration shows that the body is decelerating

speed

ratio of distance traveled to time interval

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explanation: