Question

A particle, originally at rest and located at point (3 ft, 2 ft, 5 ft), is subjected to an acceleration of a = 56t i + 12 t 2k6 ft>s2. Determine the particle’s position (x, y, z) at t = 1 s.

Answer 1

What I understand that acceleration a={56ti+12t²k}ft/s²

Answer:

The particle position is (12.333 ft, 2 ft, 6 ft )

Explanation:

Given data

Particle located at point (3 ft, 2 ft,5 ft)

Acceleration a={56ti+12t²k}ft/s²

To find

Particle position

Solution

As given acceleration is

$\int\limits^{v}_{o} \, dv=\int\limits^{t}_{o} {56ti+12t^{2}k } \, dt\\ v=28t^{2} i+4t^{3}k\\\int\limits^{r}_{r_{o} } {} \, dr=\int\limits^{t}_{o} {28t^{2} i+4t^{3}k} \, dt\\ r-r_{o}=9.333t^{3}i+t^{4}k$

r₀ is given as=(3i+2j+5k)

So

$r-r_{o}=9.333t^{3}i+t^{4}k\\r-(3i+2j+5k)=9.333t^{3}i+t^{4}k\\at\\t=1s\\r-(3i+2j+5k)=9.333(1)^{3}i+(1)^{4}k\\r-(3i+2j+5k)=9.333i+k\\r=9.333i+k+(3i+2j+5k)\\r=12.333i+2j+6k$

So the particle position is (12.333 ft, 2 ft, 6 ft )