Math

QuestionFind the distance from the origin when the particle stops, given v=9t3t2v=9t-3t^{2} m/s and starts at t=0t=0.

Studdy Solution

STEP 1

Assumptions1. The velocity of the particle is given by v=9t3tm/sv=9t-3t^{} \, m/s . At t=0t=0, the particle is at the origin3. We need to find the distance of the particle from the origin when the particle comes to instantaneous rest

STEP 2

First, we need to find the time when the particle comes to instantaneous rest. This is when the velocity is zero. We can find this by setting the velocity equation equal to zero and solving for tt.
0=9tt20 =9t -t^{2}

STEP 3

Rearrange the equation and factor out tt.
0=t(93t)0 = t(9 -3t)

STEP 4

Set each factor equal to zero and solve for tt.
t=0,93t=0t =0, \quad9 -3t =0

STEP 5

olving the second equation for tt gives ust=93=3secondst = \frac{9}{3} =3 \, seconds

STEP 6

Now that we have the time when the particle comes to rest, we can find the distance of the particle from the origin at this time. The distance can be found by integrating the velocity function from 00 to t=3t=3.
Distance=03v(t)dtDistance = \int_{0}^{3} v(t) \, dt

STEP 7

Substitute the given velocity function into the integral.
Distance=03(9t3t2)dtDistance = \int_{0}^{3} (9t -3t^{2}) \, dt

STEP 8

Calculate the integral.
Distance=[2t2t3]03Distance = \left[ \frac{}{2}t^{2} - t^{3} \right]_{0}^{3}

STEP 9

Evaluate the integral at the limits of integration.
Distance=[92(3)2(3)3][92()2()3]Distance = \left[ \frac{9}{2}(3)^{2} - (3)^{3} \right] - \left[ \frac{9}{2}()^{2} - ()^{3} \right]

STEP 10

Calculate the distance.
Distance=[92(9)27]0=40.527=13.5mDistance = \left[ \frac{9}{2}(9) -27 \right] -0 =40.5 -27 =13.5 \, mThe distance of the particle from the origin when the particle comes to instantaneous rest is13.5 meters.

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