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A car is traveling at a constant speed until its brakes are applied. The car's speed changes at a rate given by \(-0.08t \, \text{m/s}^2\) after the brakes are applied, where \(t\) (in seconds) is the time since the brakes were applied. Three seconds after the brakes are applied, the speed of the car is \(5 \, \text{m/s}\).

How far will the car travel with the brakes applied before it stops?

Answer :

The car will travel a distance of 52.08 seconds will the brakes applied before it stops

The acceleration of the car after t seconds that the brake was applied is given as:

[tex]a=-0.08t m/s^2[/tex]

The speed of the car after 3 seconds that the brake was applied, u = 5 m/s

The speed of the car when it eventually stops, v = 0m/s

3 seconds after the brake was applied:

The time, t = 3 seconds

substitute t = 3 into the acceleration [tex]a=-0.08t m/s^2[/tex]

[tex]a=-0.08(3)m/s^2\\\\a = -0.24m/s^2[/tex]

To calculate the distance traveled by the car after the brakes are applied, use the equation below

[tex]v^2=u^2+2as[/tex]

Substitute u = 5, a = -0.24, and v = 0 into the equation above in order to calculate the distance

[tex]0^2=5^2+2(-0.24)s\\\\0=25-0.48s\\\\0.48s = 25\\\\s = \frac{25}{0.48}\\\\s = 52.08 m[/tex]

The car will travel a distance of 52.08 seconds will the brakes applied before it stops

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The distance a car travels before the brakes are applied gives a reaction time of 1.5 seconds.

A car is moving at 55 mph (26 m/s) on flat ground when a hazard is spotted 85 m (278 ft) ahead.

Determine the distance the car travels before the brakes are applied considering a 1.5 s reaction time.

The car will travel 51.77 meters before the brakes are applied.