A runaway truck (truck with no brakes) traveling 45 m/s leaves the highway and enters the runaway ramp where the truck will be slowed due to soft sand that it is made of. If the sand slows the truck at -2.00 , how long,( distance), must the ramp be in order stop the truck?
V^2 = Vo^2 + 2a*d = 0
45^2 + 2*(-2)d = 0
d = ___ m.
To find the distance required to stop the truck on the runaway ramp, we can use the formula for acceleration:
acceleration = change in velocity / time
Given:
Initial velocity (u) = 45 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = -2.00 m/s²
We need to find the time it takes to decelerate the truck and bring it to a stop. Rearranging the formula above, we get:
time = (final velocity - initial velocity) / acceleration
Substituting the given values:
time = (0 - 45) / -2.00
time = -45 / -2.00
time = 22.5 seconds
Now, to find the distance covered in 22.5 seconds, we can use the formula:
distance = initial velocity * time + (1/2) * acceleration * time²
distance = 45 * 22.5 + (1/2) * -2.00 * 22.5²
distance = 1012.5 + (-1) * 2531.25
distance = 1012.5 - 2531.25
distance = -1518.75 meters
Therefore, the ramp must be at least 1518.75 meters long in order to stop the truck.
To find the distance required to stop the truck, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s - since the truck needs to stop)
u = initial velocity (45 m/s)
a = acceleration (-2.00 m/s^2)
s = distance
Substituting the known values in the equation, we have:
0^2 = 45^2 + 2(-2.00)s
Simplifying further:
0 = 2025 - 4s
Rearranging the equation to solve for s:
4s = 2025
s = 506.25 meters
Therefore, the ramp must be at least 506.25 meters long in order to stop the truck.