T-Mobile LTE 13% 7:39 p.m. a webassign.net The figure below shows a graph of v v

Question

T-Mobile LTE 13% 7:39 p.m. a webassign.net The figure below shows a graph of v versus t for the motion of a motorcyclist as he starts from rest and moves along the road in a straight line. 10 8 6 o 246 8 10 O (a) Find the average acceleration for the time interval t 0 to t-3.0 s 1.33 m/s2 (b) Estimate the time at which the acceleration has its greatest positive value. What is the value of the acceleration at that instant? 1.33 m/s2 (c) When is the acceleration zero? Acceleration is zero when t10 s and when t> 10 S. (d) Estimate the maximum negative value of the acceleration. m/s2 At what time does it occur? 10

Explanation / Answer

Part (a) average acceleration=change in velocity /time taken.

hera time taken=3s

since velocity at t=0 is 0m/s and velocity at t=3 is 4 m/s.

change in velocity=4-0=4

Average acceleration=4/3=1.33 m/s2

Part(b)Above graph is velocity versus time.

SO slope of above graph will give acceleration.We can see that at t=4 seconds slope is maximum.

Slope of a graph at a point is tangent of the angle made with positive x axis by tangent drawn at that point .

So at t=4 s acceleration is maximum.

Part(c) You have done it.

Part (d) You have done it.

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