Question:
For a scene in a movie, a stunt driver drives a 1500 kg SUV with a length of 4.25 m around a circular curve with a radius of curvature of .333 km. the vehicle is to be driven off the edge of a gully 10.0 m wide, and land on the other side 2.96 m below the initial side. What is the minimum centripetal acceleration the truck must have in going around the circular curve to clear the gully and land on the other side?
Solution:
Finding time
Δy = (1/2)gt² This equation shows the distance an object travels from rest after being released. First we have to find
2.96 = (1/2)(9.8)t² the time it took for the vertical drop by using the y equation because we know the vertical drop and
t =.777 sec gravity
Find a
14.25 = V(.777)+.5(.333x.777)² We find initial velocity first because that is where the slowest speed it has. Then we plug
V = 18.3 m/s it into the centripetal acceleration equation.
A(c) = (18.3)²/333m = 1.01 m/s²
Finding time
Δy = (1/2)gt² This equation shows the distance an object travels from rest after being released. First we have to find
2.96 = (1/2)(9.8)t² the time it took for the vertical drop by using the y equation because we know the vertical drop and
t =.777 sec gravity
Find a
14.25 = V(.777)+.5(.333x.777)² We find initial velocity first because that is where the slowest speed it has. Then we plug
V = 18.3 m/s it into the centripetal acceleration equation.
A(c) = (18.3)²/333m = 1.01 m/s²