A car is travelling at 90mph when its brakes are fully applied, producing a constant deceleration of 22 ft/sec2. Find the distance covered by the car before it comes to a stop.
(Remember: 60 mph equals 88 ft/sec.)
23
To find the distance covered by the car before it comes to a stop, we need to use the formulas of motion under constant acceleration.
First, we need to convert the initial velocity from mph to ft/sec. As stated, 60 mph equals 88 ft/sec. Therefore, the initial velocity of the car is 90 mph * (88 ft/sec / 60 mph) = 132 ft/sec.
The acceleration is given as 22 ft/sec^2. We will consider this as a negative value since it is deceleration.
We have the following known values:
Initial velocity (u) = 132 ft/sec
Acceleration (a) = -22 ft/sec^2 (negative because it's deceleration)
Final velocity (v) = 0 ft/sec (since the car comes to a stop)
We need to find the distance covered (s).
We can use the equation:
v^2 = u^2 + 2as
Rearranging the equation gives:
s = (v^2 - u^2) / (2a)
Plugging in the given values, we have:
s = (0^2 - 132^2) / (2 * -22)
s = (-17424) / (-44)
s = 396 ft
Therefore, the distance covered by the car before it comes to a stop is 396 ft.