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.
