If the position of aparticle is given by x=5.0-9.8t+6.4t^2, what is the velocity and acceleration of the particle at t=4.0s?

Mar 17, 2018

$v \left(4\right) = 41.4 \setminus \textrm{\frac{m}{s}}$
$a \left(4\right) = 12.8 \setminus {\textrm{\frac{m}{s}}}^{2}$

Explanation:

$x \left(t\right) = 5.0 - 9.8 t + 6.4 {t}^{2} \setminus \textrm{m}$

$v \left(t\right) = \frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}} = - 9.8 + 12.8 t \setminus \textrm{\frac{m}{s}}$

$a \left(t\right) = \frac{\mathrm{dv} \left(t\right)}{\mathrm{dt}} = 12.8 \setminus {\textrm{\frac{m}{s}}}^{2}$

At $t = 4$:

$v \left(4\right) = - 9.8 + 12.8 \left(4\right) = 41.4 \setminus \textrm{\frac{m}{s}}$
$a \left(4\right) = 12.8 \setminus {\textrm{\frac{m}{s}}}^{2}$

Mar 17, 2018

The given equation can be compared with $s = u t + \frac{1}{2} a {t}^{2}$

which is an equation of position-time relationship of a particle moving with constant acceleration.

So,rearranging the given equation,we get,

$x = 5 - 9.8 \cdot t + \frac{1}{2} \cdot 12.8 {t}^{2}$ (also see, at $t = 0 , x = 5$)

So,the acceleration of the particle is constant i.e $12.8 m {s}^{-} 2$ and initial velocity $u = - 9.8 m {s}^{-} 1$

Now,we can use the equation, $v = u + a t$ to find velocity after $4 s$

So, $v = - 9.8 + 12.8 \cdot 4 = 41.4 m {s}^{-} 1$