A car is travelling at 96 km/h and subjected to a deceleration of 4.9 m/s2 when the brakes are applied and the car finally comes to a stop. Determine the following:(a) the time it takes for the car to stop, t = Answer seconds,(b) the distance (in metres) the car travels in this time, s = Answer metres.
Question
A car is travelling at 96 km/h and subjected to a deceleration of 4.9 m/s2 when the brakes are applied and the car finally comes to a stop. Determine the following:(a) the time it takes for the car to stop, t = Answer seconds,(b) the distance (in metres) the car travels in this time, s = Answer metres.
Solution
To solve this problem, we first need to convert the speed from km/h to m/s because the deceleration is given in m/s².
1 km/h = 1000 m/3600 s = 0.27778 m/s
So, 96 km/h = 96 * 0.27778 = 26.67 m/s
(a) We can use the equation of motion v = u - at to find the time it takes for the car to stop. Here, v is the final velocity, u is the initial velocity, a is the acceleration (which is negative because it's deceleration), and t is the time.
Setting v = 0 (because the car comes to a stop), we get:
0 = 26.67 - 4.9t
Solving for t, we get:
t = 26.67 / 4.9 = 5.44 seconds
(b) We can use the equation of motion s = ut - 0.5at² to find the distance the car travels in this time. Here, s is the distance.
Substituting the values we have, we get:
s = 26.67 * 5.44 - 0.5 * 4.9 * (5.44)² = 144.93 - 72.47 = 72.46 metres
So, the time it takes for the car to stop is 5.44 seconds and the distance the car travels in this time is 72.46 metres.
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