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 88m. if the car continues to slows down at same rate ,aftervwhat further distance will it be further be brought to rest.
average velocity is 11 m/s ... (15 + 7) / 2
time is 8 s ... 88 / 11
acceleration is -1 m/s^2 ... (7 - 15) / 8
time to stop is 7 s ... -7 / -1
ave stopping velocity is 3.5 m/s
... (7 + 0) / 2
stop distance = (stop time) * (ave stop vel)
V^2 = Vo^2 + 2a*d.
7^2 = 15^2 + 2a*88
-176a = 15^2-7^2
a = -1.0 m/s^2.
0 = 15^2 + 2*(-1)d.
d = 112.5 m.
112.5 - 88 = 24.5 m. further.
To find out the further distance the car will travel before coming to a rest, we can use the equations of motion.
First, let's write down the given information:
Initial velocity (u) = 15 m/s
Final velocity (v) = 7 m/s
Distance (s) = 88 m
We need to calculate the acceleration (a) using the formulas of motion.
The first equation of motion is: v^2 = u^2 + 2as
Rearranging the equation to solve for acceleration:
a = (v^2 - u^2) / 2s
Now, plug the given values into the equation:
a = (7^2 - 15^2) / (2 * 88)
a = (-176) / 176
a = -1 m/s^2
We observe that the term "retardation" is actually the negative acceleration.
Now, we have the acceleration (a) as -1 m/s^2.
To find the further distance, we can use the second equation of motion: v^2 = u^2 + 2as
Since the car comes to rest (v = 0), we can plug in the values:
0 = 7^2 + 2 * (-1) * s
Rearranging the equation to solve for distance:
2s = -49
s = -24.5 m
The negative value of distance indicates that the car will move backward in the opposite direction.
Therefore, the car will travel an additional 24.5 meters in the opposite direction to come to a complete rest.