Starting from rest, you move with a constant acceleration of 1.2 m/s^2 in the po

Question

Starting from rest, you move with a constant acceleration of 1.2 m/s^2 in the positive-x direction for 12 s. At the end of the first 12 seconds, you switch to an acceleration of 0.95 m/s^2 in the negative-x direction for another 12 s. What was the maximum speed that you attained? What was the total distance that you traveled during the whole trip? What was your average speed for the trip? What was your average x-component of velocity? How do your answers to parts (c) and (d) compare? Will this always be the case? Why or why not?

Explanation / Answer

5) a) initial velocity is Vo = 0 m/sec

time of travel is t = 12 sec

accelaration is a = 1.2 m/s^2

then using

V= Vo+(a*t)

V = 0 + (1.2*12) = 14.4 m/sec

Vmax = V +(a*t) = 14.4+(0.95*12) = 25.8 m/sec is the answer for a)

for first 12 sec

S1 = (Vo*t) +(0.5*a*t^2) = 0 +(0.5*1.2*12^2) = 86.4 m

for the next 12 sec

S2 = (Vo*t)+(0.5*a*t^2) = (14.4*12)+(0.5*0.95*12^2) = 241.2 m

total distance travelled is S = S1+S2 = 86.4+241.2 = 327.6 m

c) average speed is Vavg = total distance / time taken = 327.6 / 24 = 13.65 m/sec

D) average x-component of velocity is Vavg = (86.4-241.2)/24 = -6.45 m/sec

answer is 6.45 m/sec along negative X-axis

E) average speed in C) is greater than average velocity in D)

this is not always true ,this is true only in the case ,which has both +X-axis and -X-axis ,if they travel only along one axis ,this is not true

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