A car is braked so that it slows down with a uniform retardation from 15m/s to 7m/s while it travels a distance of 88m. If a car continues to slow down at the same rate, after what further distance will it be brought to rest?
find a
average speed = (15+7)/2 = 11 m/s
so t = 88/11 = 8 seconds
so
a = (7-15)/8 = - 1 m/s^2
v = Vi + a t
0 = 7 - 1 t
t = 7 during final stop
d = Vi t + (1/2) a t^2
d = 7*7 - (1/2)49
d = 49/2 = 24.5 meters
A car travelling with a speed of 15m/s is braked and it slows down with uniform retardation.It covers of 88 m as its velocity reduces to 7m/s.If the car continues to slow down with the same ratr,further distance will it be brought to rest?
To find the distance at which the car will come to rest, we can use the equations of motion.
The equation that relates the initial velocity (u), final velocity (v), acceleration (a), and distance (d) is:
v^2 = u^2 + 2ad
We know the initial velocity (u) is 15 m/s, final velocity (v) is 7 m/s, and the distance (d) is 88 m. We also know that the car is slowing down with uniform retardation, which means the acceleration (a) will be negative.
Plugging in the values into the equation, we have:
7^2 = 15^2 + 2a(88)
49 = 225 + 176a
Rearranging the equation, we get:
176a = -176
Dividing both sides by 176:
a = -1 m/s^2
Now, to find the distance at which the car will come to rest, we need to find the distance (d) when the final velocity (v) is 0 m/s.
Plugging the values into the equation:
0^2 = 15^2 + 2(-1)d
0 = 225 - 2d
2d = 225
d = 225/2
d = 112.5 m
Therefore, the car will be brought to rest after a further distance of 112.5 meters.
To find the answer, we can use the equations of motion. The equation that relates initial velocity (u), final velocity (v), distance (s), acceleration (a), and time (t) is:
v^2 = u^2 + 2as
In this case, the initial velocity (u) is 15 m/s, the final velocity (v) is 7 m/s, and the distance (s) is 88 m. We need to find the distance (s) when the final velocity (v) is 0 m/s, which means the car is brought to rest.
Since the car continues to slow down at the same rate, the acceleration (a) remains the same. Let's assume the final distance when the car is brought to rest is d.
Initially:
u = 15 m/s
v = 7 m/s
s = 88 m
Finally:
u = 7 m/s
v = 0 m/s
s = d (unknown)
Using the equation of motion, we can substitute the initial and final values:
v^2 = u^2 + 2as
(0^2) = (7^2) + 2a(d)
0 = 49 + 2ad
Rearranging the equation:
2ad = -49
Now, we need to find the value of acceleration (a). When the car initially slows down from 15 m/s to 7 m/s over a distance of 88 m, the acceleration remains constant. We can use another equation of motion to find acceleration:
(v - u) = at
Here, v = 7 m/s, u = 15 m/s, and t is the time taken to travel a distance of 88 m. We need to find acceleration (a).
(v - u) = at
(7 - 15) = a(t)
-8 = a(t)
The time taken to cover a distance of 88 m can be calculated using the equation:
s = ut + (1/2)at^2
88 = (15)t + (1/2)(-8)t^2
Simplifying the equation:
88 = 15t - 4t^2
Rearranging the equation:
4t^2 - 15t + 88 = 0
We can solve this quadratic equation to find the value of t. By substituting this value back into the equation for acceleration, we can find the value of a.
Once we have the value of a, we can substitute it back into the equation for finding the distance (s) when the car is brought to rest:
2ad = -49
Now, solve for d by substituting the value of a obtained earlier. This will give us the distance at which the car will be brought to rest.