y Notes O Ask Your Teac 6. 0/1 points | Previous Answers SerPSE10 2.7.0P012 A mo

Question

y Notes O Ask Your Teac 6. 0/1 points | Previous Answers SerPSE10 2.7.0P012 A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 49.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h, determine its minimum stopping distance (in ft), assuming the same rate of acceleration Look carefully at the kinematic equation relating the nal speed, initial speed, acceleration, and change in position. Note that the velocities are squared. Did you remember to square the velocities? ft Need Help? Read It

Explanation / Answer

v1= 44 miph= 64.53 ft/s

s1= 49 ft

Using 3rd equation of motion

a= v1^2/(2*s1)= 64.53^2/(2*49)

a= 42.49 ft/s^2

Now

V2= 115.867 m/s

s2= v2^2/(2* 42.49)

s2= 158 ft

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Comment in case any doubt.. good luck

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