A box accidentally falls from the back of a truck and hits the ground with a speed of 15m/s. It slides along the ground for a distance of 45 m before coming to rest. Determine
a) the length of time the box slides before stopping
c) the time it takes to slide the last 10 m.
a) Its average speed while coming to a stop is 7.5 m/s. Divide 45 m by that average speed to get the time required. You get 6.00 s
b)The speed while slowing down is
V = 15 [1 - (t/6)]
The time required to go 35 m is given by
Integral V dt = 35 = 15 t - (5/4)t^2
0 to t
Solve that quadratic for t. That will be the time required to slide the first 35 m. Subtract that from 6 s for the answer.
Why is the velocity 7.5?
A better solution is using the big 5 equations
the area under the velocity versus time graph gives you the total displacement so;
30.t/2 = 45
t= 30
To determine the length of time the box slides before stopping, we need to calculate the time it takes for the box to decelerate and come to rest. We can use the equation of motion that relates initial velocity (vi), final velocity (vf), acceleration (a), and time (t):
vf = vi + at
In this case, the box starts from rest (vi = 0 m/s) and comes to rest (vf = 0 m/s), so the equation becomes:
0 = 15 m/s + a*t
Since the box comes to rest, the final velocity is zero. We can rearrange the equation to solve for time:
a*t = -15 m/s
Now we need to determine the acceleration. To do this, we can use the equation that relates acceleration, final velocity, and distance:
vf^2 = vi^2 + 2ad
Since the final velocity is zero, the equation becomes:
0 = 15^2 + 2*a*45
225 = 90a
Solving for acceleration:
a = 225 / 90
a = 2.5 m/s^2
Now we can substitute the value of acceleration back into the first equation to solve for time:
2.5 * t = -15
Solving for time:
t = -15 / 2.5
t = -6 seconds
Since time cannot be negative in this context, we disregard the negative solution. Therefore, the length of time the box slides before stopping is 6 seconds.
To determine the time it takes to slide the last 10 m, we can use the equation of motion that relates initial velocity, final velocity, acceleration, and distance:
vf^2 = vi^2 + 2ad
In this case, the initial velocity (vi) is 15 m/s, the final velocity (vf) is 0 m/s, the acceleration (a) is -2.5 m/s^2 (negative because it is decelerating), and the distance (d) is 10 m. The equation becomes:
0 = 15^2 + 2*(-2.5)*10
Solving for time:
225 = -50t^2
Rearranging the equation:
t^2 = -225 / -50
t^2 = 4.5
Taking the square root:
t = √4.5
t ≈ 2.12 seconds
Therefore, it takes approximately 2.12 seconds for the box to slide the last 10 m.