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

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SUMMARY

The average resultant force on a 3000 kg truck making a 90-degree turn can be calculated using the change in velocity and the time taken for the turn. The truck initially travels at 4.0 m/s and emerges with a speed of 7.0 m/s after 5.0 seconds. The change in velocity (ΔV) is 3.0 m/s, and using the formula F = m*a, where acceleration (a) is ΔV/Δt, the average resultant force is determined to be 1800 N.

PREREQUISITES
  • Understanding of Newton's Second Law (F = m*a)
  • Knowledge of vector quantities and their manipulation
  • Familiarity with the concept of acceleration
  • Basic principles of centripetal force
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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.
 
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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.
 

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