What is the average resultant force on a truck making a 90 degree turn?

[obaid]

Homework Statement

a 3000 kg truck traveling at a speed of 4.0 m/s makes a 90 degree turn in a time of 5.0 s and emerges from this turn with a speed of 7.0 m/s. what is the magnitude of the average resultant force on the truck during this turn?

Homework Equations

The Attempt at a Solution

I think this might be a centripetal force problem but the change in velocities is messing me up.

 

[LowlyPion]

What is force? F = m*a

What is acceleration? ΔV/Δt

So ... what is the ΔV ... keeping in mind that V is a vector?

You have the Δt conveniently given as 5 sec.

 

[nlightner]

You are correct in thinking that this is a centripetal force problem. To determine the magnitude of the average resultant force on the truck during the turn, we can use the equation F = m(v^2/r), where F is the centripetal force, m is the mass of the truck, v is the velocity, and r is the radius of the turn.

First, we need to find the radius of the turn. We can use the formula v = ωr, where ω is the angular velocity. In this case, the truck makes a 90 degree turn in 5.0 seconds, so the angular velocity would be 90 degrees/5 seconds = 18 degrees/s. Converting this to radians/s, we get ω = (18 degrees/s)(π/180 degrees) = 0.314 radians/s.

Now, we can plug in the values into the equation F = m(v^2/r). The mass of the truck is given as 3000 kg, the initial velocity is 4.0 m/s, and the final velocity is 7.0 m/s. The radius of the turn can be found using the formula v = ωr, where v is the initial velocity. So, r = v/ω = (4.0 m/s)/(0.314 radians/s) = 12.74 m.

Plugging in all the values, we get F = (3000 kg)((7.0 m/s)^2)/(12.74 m) = 3679.76 N. Therefore, the magnitude of the average resultant force on the truck during this turn is approximately 3679.76 N.