please help me with this :( wrong items at @t=1.47 x=? @t=3.02 v=? a=? The parti

Question

please help me with this :(
wrong items at
@t=1.47
x=?

@t=3.02
v=?
a=?

The particle will cross the origin when t=?

wrong item at
t=?

Return to Blackboard am, Dynamics, 7e Your answer is partially correct. Try again. A partidle starts from rest at x 3.8 m and moves along the x-axis with the velocity history shown. Plot the corresponding seconds. Find the time t when the partide crosses the arigin. After you have the plots, answer the questions. 63???? Questions At t -0.66 s,I-2.43 At t-1.47 s, At t 3.02 s, , v-4.158 3.95 m/s, “ a 10.02 The partidle will cross the origin when t- Click if you would like to Show Work for this question: LINK TO TEXT

Explanation / Answer

v = dx/dt

a = dv / dt = d2t/dt2

Between t = 0 and t = 1 sec:

dv/dt = (6.3 - 0) / (1 - 0) = 6.3 m/s^2

Integrating it with respect to time, v = 6.3t + C1

At t = 0, we have v = 0. Hence, C1 = 0

v = 6.3t

dx/dt = 6.3t

Integrating it with time, x = 6.3t^2 /2 + C2

At t = 0, we have x = -3.8. Thus C2 = -3.8

x = 3.15 t2 - 3.8

At t = 1 sec, we'll have x = 3.15* 12 - 3.8 = -0.65 m

Between t = 1 and t = 2 sec:

v = 6.3 m/s

a = dv/dt = 0

dx / dt = 6.3

Integrating it, x = 6.3t + C3

At t = 1 sec, we have x = -0.65 m. Therefore, C3 = -0.65 - 6.3*1 = -6.95

x = 6.3t - 6.95

At t = 2 sec, we'll have x = 6.3*2 - 6.95 = 5.65 m

At t > 2 sec:

dv/dt = (-1.9 - 6.3) / (4-2) = -4.1 m/s^2

Integrating it with respect to time, v = -4.1t + C4

At t = 2 sec, we have v = 6.3 m/s. Hence, C4 = 6.3 + 4.1*2 = 14.5

v = -4.1t + 14.5

dx/dt = -4.1t + 14.5

Integrating it with time, x = -4.1t^2 /2 + 14.5t + C5

At t = 2, we have x = 5.65. Thus C5 = -15.15

x = -2.05 t2 + 14.5t - 15.15

Using these equations now we can evaluate.

At t = 0.66, x = 3.15 * (0.66)2 - 3.8 = -2.42786 m

At t = 0.66, v = 6.3 * 0.66 = 4.158 m/s

At t = 0.66, a = 6.3 m/s^2

At t = 1.47, x = 6.3 * 1.47 - 6.95 = 2.311 m

At t = 1.47, v = 6.3 m/s

At t = 1.47, a = 0

At t = 3.02, x = -2.05 * (3.02)2 + 14.5*3.02 - 15.15 = 9.94318 m

At t = 3.02, v = -4.1 * 3.02 + 14.5 = 2.118 m/s

At t = 3.02, a = -4.1 m/s^2

At t = 1 sec, we found that x = -0.65 m and at t= 2 sec, x = 5.65 m. Hence, it'll cross origin (x = 0) between 1 and 2 sec.

x = 6.3t - 6.95

Putting x = 0, we get t = 1.103 sec

-------------------------

For straight portions of train, speed = 0.25 m/s and total straight distance = 2 + 2 = 4 m

Total Time taken on straight distance = 4 /0.25 = 16 s

dv/ds = (dv/dt) / (ds/dt)

Between s = 2 and 2 + pi/2,

we have dv/ds = (0.1 - 0.25) / (2+ pi/2 - 2) = -0.3 / pi

dv/dt = (-0.3 / pi) (ds/dt)

dv/dt = (-0.3 / pi) v

dv / v = (-0.3 / pi) dt

ln v = (-0.3 / pi) t + C1

At t = 8 sec, v = 0.25 m/s. Thus C1 = (ln 0.25) + (0.3*8 / pi) = -0.622

ln v = (-0.3 / pi) t - 0.622

When v = 0.1 s, we'll have t = ((ln 0.1) + 0.622) / (-0.3 / pi) = 17.59 s

Between s = 2 + pi/2 and 2 + pi,

we have dv/ds = (0.25 - 0.1) / (2+ pi - (2 + pi/2)) = 0.3 / pi

dv/dt = (0.3 / pi) (ds/dt)

dv/dt = (0.3 / pi) v

dv / v = (0.3 / pi) dt

ln v = (0.3 / pi) t + C2

At t = 17.59 sec, v = 0.1 m/s. Thus C2 = (ln 0.1) - (0.3*17.59 / pi) = -3.983

ln v = (0.3 / pi) t - 3.983

When v = 0.25 m/s, we'll have t = ((ln 0.25) + 3.983) / (0.3 / pi) = 27.179 m/s

Total time from s = 2 to s = 2 + pi will be 27.179 - 8 = 19.179 s

Total time on track is therefore 16 + 2*19.179 = 54.358 s

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