6. An object is in motion and it's distance, in meters, can...

6. An object is in motion and it's distance, in meters, can be determined by the function: $\frac{1}{3}t^3 - 3t^2 + 8t - 5$, where $t$ is in seconds. (a) Find the time when the object is at rest. (b) Find the time when the object is traveling at a constant velocity. 7. A particle is moving along a straight line where its position is found by the function: $s(t) = t^3 - 2t^2$ (a) Find the velocity and acceleration functions. (b) Find the time when the particle is moving at a constant velocity. (c) Find the time intervals when the particle is slowing down and speeding up.

Transcript

00:01 We have been given an equation the position of the particle as a function of time t, that is tq minus 70 square, plus 11 t minus 2 .5.

00:15 Now a velocity is given up by d s upon d t.

00:20 So upon differentiation of this equation, we will get 3 t squared minus 14 t plus 11 as the velocity.

00:32 Of the particle and acceleration will be given as again differentiating dv by d t or d2 s upon d t square that is 60 minus 14 so this is the part a now part b the particle will be raised will be raised when velocity will be zero and it will be going to speed up when acceleration will be positive and it will go into slow down when acceleration will be negative that is velocity is decreasing here and velocity is increasing when the body is speeding up so acceleration automatically comes out to positive so now on solving v equals to 0 or we have got v equals to 3 t squared minus 14 t plus 11 equals to 0 so it can be written at 3 t2 minus 3 t plus 11 equals to 0 now taking t common it will be 3 t minus 11 minus 1 3 t minus 11 equals to 0 so t comes out if it could be 11 by 3 or t comes equal to 1 that means at t equals to 11 by 3 second or t equals to 1 second the velocity will be equal to 0 now when acceleration will be positive, acceleration will be positive or the velocity of the body is going to speed up.

02:26 I'm solving here, now putting the equation of the acceleration that is 60 minus 14 must be greater than 0.

02:36 That comes equal to 14 by 6.

02:41 And acceleration will be negative.

02:44 That is body is going to slow down and 60 minus 14 less than 0.

02:50 That is t is less than 14 upon 6 so the concept here is if we take this line and draw the time scale at t equals to 0 t equals to 1 now let here it equals to 14 upon 6 so at rightwards the velocity of the body is going to speed up and left towards the velocity of the body is going to slow down that means at all time t greater than 14 by 6 seconds the body is going to speed up and t less than 6 second let's suppose t equals 2 second the velocity of the body will go into slow down.

03:38 Now third part c the values where the particle change direction it will change when the velocity becomes positive to negative or negative positive.

03:59 So all we require to solve the velocity to 0 and plot the graph.

04:07 Now velocity is given as 3 t squared minus 14 t plus 11.

04:15 Now on solving we'll get t equals to 1 and t equals to 14 upon 3.

04:23 No, it's 11 upon 3 seconds...

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