Question

The acceleration of a particle as it moves along a straight line is given by a = (2t – 1) m/s2. If s = 1 m and v = 2 m/s when t = 0, determine the particle’s velocity and position when t = 6 s. Also determine the total distance the particle travels during this time period.

Answer 1

1) Velocity at t =6 =32m/s

2) Position of particle at t = 6 secs = 67 meters

3) Distance covered in 6 seconds equals 72 meters.

Step-by-step explanation:

By definition of acceleration we have

`a=(dv)/(dt)\\\\dv=a(t)dt\\\\\int dv=\int a(t)dt\\\\v(t)=\int (2t-1)dt\\\\v(t)=t^(2)-t+c_(1)`

Now at t = 0 v = 2 m/s thus the value of constant is obtained as

`2=0-0+c_(1)\\\\\therefore c_(1)=2`

thus velocity as a function of time is given by

`v(t)=t^(2)-t+2`

Similarly position can be found by

`x(t)=\int v(t)dt\\\\x(t)=\int (t^(2)-t+2)dt\\\\x(t)=(t^(3))/(3)-(t^(2))/(2)+2t+c_(2)`

The value of constant can be obtained by noting that at time t = 0 x = 1.

Thus we get

`1=0+0+0+c_(2)\\\\\therefore c_(2)=1`

thus position as a function of time is given by

`x(t)=(t^(3))/(3)-(t^(2))/(2)+2t+1`

Thus at t = 6 seconds we have

`v(6)=6^(2)-6+2=32m/s`

`x(6)=(6^(3))/(3)-(6^(2))/(2)+2* 6 +1=67m`

The path length can be obtained by evaluating the integral

`s=\int_(0)^(6)\sqrt{1+(t^(2)-t+2)^(2)}\cdot dt\\\\s=72meters`