A car braked at a constant deceleration of 45 ft / s2, producing a skid mark measuring 115 feet before coming to a stop.
How fast was the car travelling when the brakes were first applied?
To find the initial velocity (speed) of the car when the brakes were first applied, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 ft/s as the car comes to a stop)
u = initial velocity (unknown)
a = acceleration (deceleration due to braking, -45 ft/s^2 as it is deceleration)
s = distance traveled (skid mark length, 115 ft)
Plugging in the given values, we have:
0 = u^2 + 2(-45)(115)
Simplifying, we get:
0 = u^2 - 10350
Rearranging the equation, we have:
u^2 = 10350
Taking the square root of both sides, we find:
u = ± √10350
Since the initial velocity can only be positive, we can ignore the negative sign.
Therefore, the initial velocity of the car when the brakes were first applied was approximately u = √10350 ft/s.
To find the speed at which the car was traveling when the brakes were first applied, you can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 ft/s since the car comes to a stop)
u = initial velocity (unknown)
a = acceleration (deceleration, given as -45 ft/s^2)
s = distance covered (skid mark, 115 ft)
Rearranging the equation, we get:
u^2 = v^2 - 2as
Substituting the known values, we have:
u^2 = 0 - 2(-45)(115)
u^2 = 0 + 2*45*115
u^2 = 2*45*115
u^2 = 10350
Taking the square root of both sides, we find:
u = √10350
u ≈ 101.7 ft/s
Therefore, the car was traveling at approximately 101.7 ft/s when the brakes were first applied.