A car starts a drag race from rest and travels in a straight path for 1/4 mile (

Question

A car starts a drag race from rest and travels in a straight path for 1/4 mile (402 m), in a time of 5.27 s. Remember V_avg notequalto V_r (a) Assuming the cars acceleration is constant, calculate the magnitude of the acceleration in m/s^2. (b) Assuming the acceleration due to gravity is g = 9.80m/s^2 (i.e., define a new unit of acceleration, the "g" as 1.00 "g" = 9.80 m/s^2): how many g's of acceleration did the race car achieve? (c) Again assuming constant acceleration: calculate the car's speed at the end of the 402 m drag strip. Note that 2.23694 mi/hr = 1.00 m/s

Explanation / Answer

part a)

USing Newton's second law of motion

s = ut + 1/2at^2 (where u = initial velocity, t = time, a acceleration and s = distance covered)

Here car starts from rest thus u = 0m/s

402 = 1/2*a*(5.27)2

acceleration, a = 2*402/(5.27)2 = 28.95 m/s^2

part b)

1g = 9.8m/s^2

1m/s^2 = 1/9.8 g

28.95 m/s^2 = 28.95/9.8 g = 2.95 g

Thus car has achieved 2.95g of acceleration

part c)

By newton's third law we have

v^2 - u^2 = 2as (where v = final velocity, u = initial velo, a = acceleration and s = distance covered)

v^2 = 2*28.95*402

v = sqrt(2*28.95*402) = 152.56 m/s

Thus final velocity of car is 152.56 m/s

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