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An automobile passes a point PP at a speed of 84 mph. At PP it begins to decelerate at a rate proportional to time. If after 6.3 sec the car has slowed to 45 mph, what distance has it traveled in ft

Respuesta :

Answer:

Distance in ft = 597.57 feet

Step-by-step explanation:

As per the

First equation of motion

84 mph = 37.5514 m/s

45 mph = 20.1168 m/s

v-u = a* t

Substituting the given values, we get -

37.5514 m/s - 20.1168 m/s = a * 6.3

a = 2.76 m/s^2

Now as per Newton's third law of motion

v^2-u^2 = 2 *a*S

Substituting the given values, we get -

37.5514^2 - 20.1168^2 = 2 * 2.76 m/s^2 * S

S = 182.142 Meters

Distance in ft = 597.57 feet

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