A particle starts from the origin at t = 0 and moves along the positive x axis.

Question

A particle starts from the origin at t = 0 and moves along the positive x axis. A graph of the velocity of the particle as a function of the time is shown in the figure; the v_s = 7.0 m/s. What is the coordinate of the particle at t = 5.0 s? What is the velocity of the particle at t = 5.0 s? What is the acceleration of axis scale is set by vs the particle at t = 5.0 s? What is the average velocity of the particle between t = 1.0 s and t = 5.0 s? What is the average acceleration of the particle between t = 1.0 s and t = 5.0 s?

Explanation / Answer

we know, displacement = area under v-t curve

a)

at t = 5.0s,

x = (1/2)*2*7 + (4-2)*7 + (1/2)*2*7 - (1/2)*1*3.5

= 26.25 m

b) at t = 5.0s,

v = 3.5 m/s

c) at t =5.0s,

a = (v2 - v1)/(t2-t1)

= (0 - 7)/(6-4)

= -3.5 m/s^2

d) at t = 1s,

x1 = (1/2)*1*3.5 = 3.5 m

at t = 5s,
x2 = 26.25 m

so, v_avg = (x2 - x1)/(t2-t1)

= (26.25 - 3.5)/(5-1)

= 5.7 m/s

e) a_avg = (v2 - v1)/(t2-t1)

= (3.5 - 3.5)/(5-1)

= 0

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