The velocity of a particle travelling in a straight line is given by v=(6t-3t^2)

Question

The velocity of a particle travelling in a straight line is given by v=(6t-3t^2)m/s, where t is in seconds. Suppose that s=0 when t=0. Determine the particles deceleration and position at t=3.1s. How far has the particle travelled during the 3.1s time interval? What is the average speed of the particle for the time period?

I got A to be -12.6 m/s^2 and B to be -0.961 m. I can't get the distance travelled during the time interval. I tried integrating from 0 to 3.1 to get 0.961, but it's incorrect. lul

Explanation / Answer

v=(6t-3t^2)m/s

intergrating value

v = (6t 3t²) m/s, but v = ds/dt. Thus,
s = (6t 3t²) dt
s = 3t² t³ + c. When t = 0, s = 0, which implies c = 0. Hence,
s = 3t² t³.

Also, a = dv/dt = 6 6t.

therefore a=-12.6 =12.6 m/s²
Thus, at t = 3.1s,
s = 3(3.1²) 3.1³ = o m, and a = 12.6 m/s²
Therefore, we can say that the particle is at the initial position and decelerate at 12.6 m/s².(answer a)

Now, to find the distance traveled until 3.1s, simply we have to find the area of v = (6t 3t²) m/s and the x-axis
(Our x-axis here is the time traveled).
But v = 0 at t = 0 and 2, thus :

s = (6t 3t²) dt, where t is in interval [0, 2] and [2, 3.1]. Since at t = [2, 3.1], our s is in minus position, then :
= { [3t² t³] at t = [0, 2] } { [3t² t³] at t = [2, 3.1] } ,

=(0-4)-(-4-.961)
= 0-4+4-0.96
distance, s = 0.961 m

Thus, total distance traveled is 0.961 m.(answer b)
This will yielding the average speed = total distance/time traveled, which is :
Average speed = (0.961/3.1) m/s
= 0.31 m/s(

Calculation may be wrong, but concept is correct.

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