A car is braked so that it slows down with uniform retardation from 15 m/s to 7 m/s while it travels a distance of 88 m. If the car continues to slow down at the same rate, after what further distance will it be brought to rest?
in my book answer is 24.5
He travel at this speed far 4second. Brake are then apply and the body come to rest with uniform retardation in a further second graph of speed against time use the fine car acceleration
its wrong
To find the distance at which the car will be brought to rest, we can use the equations of motion.
The first equation we can use is the formula for velocity (v) as a function of initial velocity (u), acceleration (a), and distance (s):
v^2 = u^2 + 2as
Since the initial velocity (u) is 15 m/s, the final velocity (v) is 0 m/s because the car will be brought to rest, and the acceleration (a) is the same for both cases, we can rewrite the equation as:
0^2 = 15^2 + 2a(s1) [Equation 1]
Here, s1 is the distance the car travels to slow down from 15 m/s to 7 m/s, which is given as 88 m in the question.
Now, we can use the same equation to find the distance at which the car will be brought to rest. Let's call this distance s2. Since the final velocity (v) is 0 m/s, the equation becomes:
0^2 = 7^2 + 2a(s2) [Equation 2]
We can now solve equations 1 and 2 simultaneously to find the value of s2.
Equation 1: 0 = 225 + 2a(88)
Equation 2: 0 = 49 + 2a(s2)
Simplifying equation 1:
200a = -225
a = -225/200
a = -1.125 m/s^2
Substituting the value of a in equation 2:
0 = 49 + 2(-1.125)(s2)
0 = 49 - 2.25s2
2.25s2 = 49
s2 = 49/2.25
s2 ≈ 21.78 m
Therefore, the car will be brought to rest after approximately 21.78 meters.
d1 = 88 m.
V^2 = Vo^2 + 2a*d1.
a = (V^2-Vo^2)/2d1=(7^2-15^2)/176 = -1.0 m/s^2.
d2 = (V^2-Vo^2)/2a=(0-15^2)/-1.0 = 225 m
d = d2-d1 = 225 - 88 = 137 m.