Question

For a scene in a movie, a stunt driver drives a 1700 kg pickup truck with a length of 4.45 m around a circular curve with a radius 0f 0.355 km. The truck is to complete a semicircle, jump across a gully, and land on the other side 3.00 m below and 10.5 m away. What is the minimum centripetal acceleration the truck must have going around the circular curve so that the entire truck will clear the gully and land on the other side

Answer 1

`0.51\ \text{m/s}^2`

Step-by-step explanation:

t = Time taken

g = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

r = Radius of track = 0.355 km

Displacement in
`y` direction is 3 m

`y=ut+(1)/(2)gt^2\\\Rightarrow 3=0+(1)/(2)* 9.81 t^2\\\Rightarrow t=\sqrt{(3* 2)/(9.81)}\\\Rightarrow t=0.782\ \text{s}`

Displacement in
`x` direction

`x=10.5\ \text{m}`

`v=(x)/(t)\\\Rightarrow v=(10.5)/(0.782)\\\Rightarrow v=13.43\ \text{m/s}`

Centripetal acceleration is given by

`a_c=(v^2)/(r)\\\Rightarrow a_c=(13.43^2)/(355)\\\Rightarrow a_c=0.51\ \text{m/s}^2`

The minimum centripetal acceleration the truck must have is
`0.51\ \text{m/s}^2`