Question

A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping distance of the car?

Answer 1

v² - u² = 2 a ∆x

where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆x

∆x = (27 m/s)² / (16 m/s²)

∆x ≈ 45.6 m

Answer 2

The stopping distance of car achieved during the braking is of 45.56 m.

Given data:

The initial speed of car is, u = 27 m/s.

The final speed of car is, v = 0 m/s. (Because car comes to stop finally)

The magnitude of deacceleration is, $a = 8\;\rm m/s^{2}$.

In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.

Therefore,

$v^{2}=u^{2}+2(-a)s$

Here, s is the stopping distance.

Solving as,

$0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m$

Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.

Learn more about the kinematic equation of motion here:

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