Question

a cyclist who starts from the top of the hill accelerates uniformly with 0.5m/s² to reach the foot of the foot with a velocity of 54km/h. find the velocities of the cyclist at the end of 5s​

Answer 1

Explanation:

Solution



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Net acceleration is due to braking and centripetal acceleration

Due to Braking,

aT=0.5m/s2

Speed of the cyclist, v=27km/h=7.5m/s

Radius of the circular turn, r=80m

Centripetal acceleration is given as:

ac=rV2

  =(7.52)/80=0.70m/s2

Since the angle between ac and aT is 900, the resultant acceleration a is given by:

a=(ac2+aT2)1/2

a=(0.72+0.52)1/2=0.86m/s2

tanθ=aTac

where θ is the angle of the resultant with the direction of the velocity. 

tanθ=0.50.7=1.4

θ=tan−1(1.4)=54.560