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

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.

(a) Find the average acceleration for the time interval

t = 0

to

t = 6.0 s.

m/s2

(b) 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

(c) When is the acceleration zero?

Acceleration is zero when t =  s and when t >  s.

(d) Estimate the maximum negative value of the acceleration.
m/s2

At what time does it occur?
s

Explanation / Answer

(a) avg acceleration=change in velocity/time

=8/6=1.33 m/s^2

(b) greatest positive value. means optimum,the highest positive slope in the curve

It is at 4 seconds

again a=V4-0/4

=3/2=1.5 m/s^2

(c)acceleration zero at zero seconds.

Optimum value of v

(d) Highest negative slope

occurs at 8 s

it is -1 m/s^2

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