Question

At t=0, a particle is located at x=25m and has a velocity of 15ms in the positive x direction. The acceleration of the particle varies with time as shown in the diagram. What is the position of the particle at t=5.0s?

Answer 1

$\text { The position of the particle at } t=5 \text { is } \frac{25}{6} \mathrm{~m}$

Explanation:

By using the diagram given below. We have,

$a(t)=a_{i}-k \cdot t$

$k=\tan \alpha=\frac{6}{6}=1$

$a(t)=6-t$

$v(t)=\int(6-t) dt$

$v(t)=6 t-\frac{1}{2} t^{2}+C$

$\text { for } \mathrm{t}=0 \mathrm {v}=15 \mathrm{~m} / \mathrm{s} \quad \mathrm{C}=15$

$v(t)=0 t-\overline{2} t^{-}+10$

$x(t)=\int v(t) d t=\int\left(6 t-\frac{1}{2} t^{2}+15\right) d t$

$x(t)=\frac{1}{2} \cdot 6 \cdot t^{2}-\frac{1}{2} \cdot \frac{1}{3} \cdot t^{3}+15 t+C$

$x(t)=3 t^{2}-\frac{1}{6} t^{3}+15 t+C$

$f \text { or } t=0 x=25 m, C=25$

$x(b)-b c-\overline{6} x+10 t+20$

$x(5)=3 \cdot 5^{2}-\frac{1}{6} \cdot 5^{3}+15 \cdot 5+25$

$x(5)=25-\frac{125}{6}$

$x(5)=\frac{150-125}{6}$

$x(5)=\frac{25}{6}$