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 m/s^2, how long,( distance), must the ramp be in order stop the truck
v = 45-2t
truck stops at t=22.5
s = 45t - t^2
now evaluate s(22.5)
the stopping time is ... (45 m/s) / (2.00 m/s^2)
the average velocity is ... [(45 m/s) + (0 m/s)] / 2
stopping distance = stopping time * average velocity
To determine the distance, or length, that the runway ramp must be in order to stop the truck, you can use the basic kinematic equation which relates distance, time, initial velocity, and acceleration:
d = (v_f^2 - v_i^2) / (2 * a)
Where:
- d is the distance
- v_f is the final velocity (which is 0 in this case, as the truck needs to stop)
- v_i is the initial velocity of the truck (which is 45 m/s)
- a is the acceleration acting on the truck (which is -2.00 m/s^2, as stated in the problem)
Substituting the known values into the equation:
0 = (v_f^2 - (45)^2) / (2 * -2.00)
To simplify the equation, we can rearrange it to solve for v_f:
v_f = √(v_i^2 - 2ad)
Now, substitute the remaining values and solve for v_f:
v_f = √((45)^2 - 2 * -2.00 * d)
Since v_f is 0, we can write the equation as:
√(2025 + 4d) = 0
To satisfy this equation, the term inside the square root must be zero:
2025 + 4d = 0
Solving for d:
4d = -2025
Divide both sides by 4:
d = -2025 / 4
The distance (d) would be -506.25 m. However, since distance cannot be negative in this context, we conclude that there is an error in the setup of the problem or in the given value for acceleration. Please check the problem and values provided again to find the correct solution.