What Minimum Deceleration Is Required for a Car to Stop on a 35 Meter Shoulder?

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To stop a car traveling at 35 mi/h on a 35-meter shoulder, it must decelerate at a minimum rate of over 7 m/s². The initial speed converts to approximately 26.25 km/h, leading to an average speed of 13.125 m/s during deceleration. The car will cover the stopping distance in less than 5 seconds. A useful formula for calculating deceleration is v² = u² + 2ad, where v is the final velocity, u is the initial velocity, a is acceleration, and d is distance. Understanding these calculations is essential for solving similar physics problems.
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Hey just starting physics and i am lost. This is the problem:
A car traveling 35 mi/h is to stop on a 35 meter long shoulder of the road. What minimum deceleration is required?

If you could type out How you would this is would be GREATLY appreciated.

Thanks
 
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okay well 1st of all convert mi/h to km/h because the stopping area is given in metres.

The question essentially asks how rapidly must the car decelerate in order to reach a velocity of 0 over 35 metres.

Well the average speed the car will have is 1/2 its initial velocity if we take decleration to be uniform. Therefore, using 1 mi = 1.5 k, the car will have an average speed of 26.25 km/h.

This is equivalent to 437.5m/minute or 7.29166m/second.

Therefore the car will cover the 35 metres in <5seconds.

Therefore the car must decelerate at >7m/s to stop in time
 
You might also find this formula to be very useful
v_2^2 = v_1^2 + 2ad
 

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