A body which uniformly come to rest in 5.0s what is it initial volecity if the body is traveling 10m
To determine the initial velocity of a body that uniformly (meaning at a constant rate) comes to rest over a distance of 10 meters in 5.0 seconds, we can use the equations of uniformly accelerated motion.
Since the body uniformly comes to rest, the acceleration, a, is uniform and we can use the following equation to relate distance (s), initial velocity (v_0), time (t), and acceleration (a):
s = v_0 * t + (1/2) * a * t^2
Since the body comes to rest, the final velocity (v) will be 0 m/s. We are also given that:
s = 10 m (the distance over which the body comes to rest)
t = 5.0 s (the time to come to rest)
We can also rearrange the formula for final velocity with respect to time which is given by:
v = v_0 + a * t
Since the final velocity v is 0 (the body comes to rest), we can solve for acceleration a:
0 = v_0 + a * 5.0
a = -v_0 / 5.0
We need to find the initial velocity v_0 using the first equation (for distance):
10 m = v_0 * 5.0 s + (1/2) * (-v_0 / 5.0) * (5.0 s) ^ 2
Now we solve for v_0:
10 m = 5.0 * v_0 - (1/2) * (v_0 / 5.0) * 25
10 m = 5v_0 - (1/2) * (5 * v_0)
10 m = 5v_0 - (5/2) * v_0
10 m = (10/2) * v_0 - (5/2) * v_0
10 m = (5/2) * v_0
Now we isolate v_0:
v_0 = (10 m) / (5/2)
v_0 = (10 m) * (2/5)
v_0 = 20 m / 5
v_0 = 4 m/s
The initial velocity of the body is 4 meters per second.