the function s=-t^3+3t^2-3t, 0<=t<=5, gives the position of a body moving on a c

Question

the function s=-t^3+3t^2-3t,   0<=t<=5, gives the position of a body moving on a coordinate line, with s in meters and t in seconds.

a. find the bodys displacement and average velocity for the given time interval

b. find the bodys speed and acceleration at the endpoints of the interval

c. when, if ever, during the interval does the body change direction?

what is the bodys displacement fo the given time interval?     ? m

what is the average velocity for the given time interval?    ? m/s

what is the bodys speed at t=0?   ? m/s

what is the bodys speed at t=5?   ? m/s

what is the bodys acceleration at t=0?  ? m/s^2

what is the bodyys acceleration at t=5?   ? m/s^2

when, if ever, during the interval does the body change direction? select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. t=? sec

b. there is no solution

Explanation / Answer

Displacement is difference in position:

S = -t^3 + 3t^2 - 3t

s(0) = 0

s(5) = -125 + 75 - 15 = -65

Displacement = 65 m

Average velocity = diplacement / time = 65 / 5 = 13 m/s

veocity function is given by :

s ' (t) = -3t^2 + 6t - 3

v(t) = -3t^2 + 6t - 3

v(0) = -3 m/s

v(5) = -75 + 30 - 3 = -48 m/s

acceleration is

s '' (t) = v ' (t) = -6t + 6

a(t) = -6t + 6

a(0) = 6 m /s^2

a(5) = -24 m/ s^2

The body changes direction when velocity changes it signs from negative to positive.

At that time v(t) = 0

-3t^2 + 6t - 3 = 0

t^2 - 2t + 1 = 0

t = 1 sec

Body changes direction when t = 1 sec

Hope this helps. Please rate it ASAP. Thanks.

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