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    A fairground ride ends with the car moving up a slope of 20 degrees from the horizontal. The component of the weight acting parallel to the ramp is about 2.3kN. The mass of a fully loaded car is 690 kg and on the slope will decelerate the car at about 3.3ms^-2.

    If the car enters the ramp at 22ms^-1. What is the minimum length the ramp needs to be for the car to stop completely (neglect length of car).

    The ride owner decides to use a shorter ramp and installs breaks that provides an additional decelerating force of 4600N. Calculate the new stopping time.

    For the first question i got 73m but I'm really confused about the second question. Can somebody help me and explain the steps? Thanks!
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    (Original post by Smile Generator)
    A fairground ride ends with the car moving up a slope of 20 degrees from the horizontal. The component of the weight acting parallel to the ramp is about 2.3kN. The mass of a fully loaded car is 690 kg and on the slope will decelerate the car at about 3.3ms^-2.

    If the car enters the ramp at 22ms^-1. What is the minimum length the ramp needs to be for the car to stop completely (neglect length of car).

    The ride owner decides to use a shorter ramp and installs breaks that provides an additional decelerating force of 4600N. Calculate the new stopping time.

    For the first question i got 73m but I'm really confused about the second question. Can somebody help me and explain the steps? Thanks!

    The original 'decelerating force' was just the compoment of the cars weight acting parallel to the slope (2.3kN) but in the second part of the question, we now have an additional force of 4600N, so we should add these together, and then calculate the new deceleration using F=ma. Then use suvat to find the time it takes to stop.
 
 
 
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Updated: November 6, 2015
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