The figure below shows a graph of vx versus t for the motion of a motorcyclist a

Question

The figure below shows a graph of vx versus t for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line. Find the average acceleration for the time interval t = 0 to t = 4.0 s. m/s2 Estimate the time at which the acceleration has its greatest positive value. s What is the value of the acceleration at that instant? m/s2 When is the acceleration zero? Acceleration is zero when t = s and when t > s. Estimate the maximum negative value of the acceleration. m/s2 At what time does it occur? s

Explanation / Answer

a)average acceleration =v/t=v2-v1/t2-t1=6-0/4-0=6/4=1.5m/s^2

b)at t=3s    curve is more linear upward curvature        a=v/t=4/3=1.33m/s^2

c)a=0 at t=8s and t >10s as slope of the v-t curve is zero

d)a=-6/8=-0.75m/s^2 downward curvature and more linear at t=8s

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