Question

If the car's speed decreases at a constant rate from 77 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 1

The acceleration of the car,

$a = \frac{v-u}{t}$

Here, v is final velocity, u is initial velocity and t is time taken by the car.

Given $u =77 \ mil/h$ ,$v = 50 mi/h$ and $t = 3.0 s = 3.0 \times \frac{1 \ h}{3600} = 8.3 \times 10^{-4} h$

Therefore, from above equation

$a =  \frac{50 \ mi/h -77 \ mi/h}{8.33 \times 10^{-4} h} = - \frac{27 \ mi/h }{8.33 \times 10^{-4} h} = - 3.2 \times 10^{4} \ mi/h^2$.

Here, negative sign shows deceleration of a car.

Thus the the magnitude of car acceleration is $3.2 \times 10^{4} \ mi/h^2$.