Question

The driver of a car traveling on the highway suddenly slams on the brakes because of a slowdown in traffic ahead. A) If the car’s speed decreases at a constant rate from 74 mi/h to 50 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line?Answer is in mi/h^2B) What distance does the car travel during the braking period?Answer is in ft

Answer 1

Answer:

Acceleration is -30000 mi/h²

Distance travelled in the 3 seconds of deceleration is 261.888 feet

Explanation:

t = Time taken for the car to slow down = 3 s =

u = Initial velocity = 75 mi/h

v = Final velocity = 50 mi/h

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{50-75}{\frac{3}{3600}}\\\Rightarrow a=-30000\ mi/h^2$

Acceleration is -30000 mi/h²

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{50^2-75^2}{2\times -30000}\\\Rightarrow s=0.0496\ mi$

Converting to feet

1 mile = 5280 feet

0.0496 mile = 0.0496×5280 = 261.888 feet

Distance travelled in the 3 seconds of deceleration is 261.888 feet