# At t=0, a particle is located at x=25m and has a velocity of 15m/s 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?

May 27, 2016

$\text{the position of the particle at t=5 is "25/6" m}$

#### Explanation:

$\text{using diagram}$
$a \left(t\right) = {a}_{i} - k \cdot t$

${a}_{i} = 6$
$k = \tan \alpha = \frac{6}{6} = 1$

$a \left(t\right) = 6 - t$

$v \left(t\right) = \int \left(6 - t\right) d t$

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

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

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

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

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

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

"for t=0 x=25 m, C=25

$x \left(t\right) = 3 {t}^{2} - \frac{1}{6} {x}^{3} + 15 t + 25$

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

$x \left(5\right) = 75 - \frac{125}{6} + 75 + 25$

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

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

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