Question

A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position is x0 = 1.3 m at t0 = 0 s

Answer 1

Follows are the solution to this question:

Step-by-step explanation:

In point a:

Place of particles

`X(t)=\int V_(x)(t)dt`

`=\int 2t^(2)dt\\\\=(2)/(3)t^(3)+C`

`\to t=0\\\\ \to X(0)=2.3 \ m`

`\to X(0)=0+C\\\\ \to C=2.3\ m`

`\to X(t)=( (2)/(3))t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( (2)/(3))* 2.2^3 +2.3 \\\\`

`= (2)/(3)* 10.648 +2.3\\\\= (21.296)/(3)+2.3\\\\ = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m`

In point b:

when
`t=2.2 \ s`

the Particle velocity
`(V)=2 * 2.22 =9.68\ (m)/(s)`

In point c:

Calculating the Particle acceleration:

`\to a=(dV)/(dt) =4\ t\\\\\to t=2.2 \ s\\\\\to a=4* 2.2 =8.8 \ (m)/(s^2)`