Help with this physics problem

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In summary, the question asks for the minimum deceleration required for a car traveling at 35 mi/h to come to a stop on a 35 meter shoulder of the road. To solve this, the average speed of the car is calculated to be 26.25 km/h, which is equivalent to 7.29166 m/s. Using the formula v_2^2 = v_1^2 + 2ad, it is determined that the car must decelerate at a rate of >7 m/s to stop in time.
  • #1
Janzen
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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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  • #2
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
 
  • #3
You might also find this formula to be very useful
[tex]v_2^2 = v_1^2 + 2ad[/tex]
 

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