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

Answer:

Follows are the solution to this question:

Explanation:

In point a:

Place of particles

$X(t)=\int V_{x}(t)dt$

       $=\int 2t^{2}dt\\\\=\frac{2}{3}t^{3}+C$

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

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

$\to X(t)=( \frac{2}{3})t^3 + 2.3\\\\ \to t=2.2\\\\\to X=( \frac{2}{3})\times 2.2^3 +2.3$

        $= \frac{2}{3}\times 10.648 +2.3\\\\= \frac{21.296}{3}+2.3\\\\ = 7.09+2.3\\\\ =9.39\\\\ =9.4\ m$

In point b:

when $t=2.2 \ s$

the Particle velocity  $(V)=2 \times 2.22 =9.68\ \frac{m}{s}$

In point c:

Calculating the Particle acceleration:

$\to a=\frac{dV}{dt} =4\ t\\\\\to t=2.2 \ s\\\\\to a=4\times 2.2 =8.8 \ \frac{m}{s^2}$