Question

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 48 ft/s2. What is the distance covered before the car comes to a stop

Answer 1

Answer:

Distance covered by the car is 56.01 feet.

Explanation:

It is given that,

Initial velocity, u = 50 mi/h = 73.33 ft/s

Constant deceleration of the car, $a=-48\ ft/s^2$

Final velocity, v = 0

Let s is the distance covered before the car comes to rest. It can be calculated using third equation of motion as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{0^2-(73.33\ ft/s^2)^2}{2\times -48\ ft/s^2}$

s = 56.01 ft

So, the distance covered before the car comes to a stop is 56.01 feet. Hence, this is the required solution.

Answer 2

The distance covered by the car as it comes to rest is 17.1 m.

To calculate the distance covered before the car come to stop, we use the formula below.

Formula:

• v² = u²+2as............... Equation 1

Where:

• s = distance covered by the car
• v = final velocity
• u = initial velocity
• a = acceleration of the car

make s the subject of the equation

• s = (v²-u²)/2a........................Equation 2

From the question,

Given:

• v = 50 mi/h = (50×0.447) = 22.35 m/s
• u = 0 m/s
• a = 48 ft/s² = (48×0.3048) = 14.63 m/s²

Substitute these values into equation 2

• s = (22.35²-0²)/(2×14.63)
• s = 17.1 m

Hence, the distance covered by the car as it comes to rest is 17.1 m.

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