In the graph given, the initial velocity (VO) 5 31.2 m/s (in positive x directio

Question

In the graph given, the initial velocity (VO) 5 31.2 m/s (in positive x direction). If t1 = 11.2 seconds, t2 is unknown, and the average velocity for the entire trip is 21.7 m/s in the positive x direction, what is the acceleration of the object from t1 to t2 in m/s^2? If the acceleration is in the negative x direction, include a negative sign, but do not include a positive sign for positive answers. (Hint write down the general equation for average velocity and then using the graph (area), write down an equation for the total displacement of the trip. This can be used to find t2 and from there and that can be used to find the acceleration from the graph.)

Explanation / Answer

here,

the initial velocity is , u = 31.2 m/s

t1 = 11.2 s

average velocity , vavg = 21.7 m/s

let the accelration be a

average velocity = total distance / total time taken

21.7 = u*t1 +(u*t2 + 0.5 * a*t2^2) /(t1 + t2)

21.7 = (31.2 * 11.2 + 31.2 *(t2 - 11.2) + 0.5 * a * (t2 - 11.2)^2)/(t2) ...(1)

and ,using first equation of motion

0 = 31.2 + a * (t2 - 11.2) ...(2)

from equation (1) and (2)

a = -1.79 m/s^2

the accelration of the object from t1 to t2 is -1.79 m/s^2

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