Question

A truck covers 40.0 m in 8.5 s while smoothly accelerating to a final velocity of 2.8 m/s. What was the original velocity? What is the acceleration?

Answer 1

Task. A truck covers 40.0 m in 8.5 s while smoothly accelerating to a final velocity of 2.8 m/s. What was the original velocity, $v_{0}$? What is the acceleration, $a$?

Solution. Assume that the acceleration of the truck is constant. Then the distance $d(t)$ covered by truck at time $t$, and the velocity $v(t)$ at the same time are given by the following formulas:

$d(t)=v_{0}t+\frac{at^{2}}{2},$

$v(t)=v_{0}+at.$

By assumption,

$d(8.5\ s)=40.0\ m,$

and

$v(8.5\ s)=2.8\ m/s.$

Denote

$t=8.5\ s,$

$d_{1}=d(t_{1})=d(8.5\ s)=40.0\ m,\qquad v_{1}=v(t_{1})=v(8.5\ s)=2.8\ m/s.$

Then we obtain the following system of equations:

$v_{0}t+\frac{at^{2}}{2}=d_{1},\qquad v_{0}+at=v_{1}.$

From the second equation we get:

$at=v_{1}-v_{0}.$

Substituting this formula into the first equation we obtain

$v_{0}t+\frac{(v_{1}-v_{0})t}{2}=d_{1},$

whence

$\frac{2v_{0}+v_{1}-v_{0}}{2}\cdot t=d_{1},$

$\frac{v_{0}+v_{1}}{2}=\frac{d_{1}}{t},$

$v_{0}=\frac{2d_{1}}{t}-v_{1},$

and therefore

$a=\frac{v_{1}-v_{0}}{t}=\frac{v_{1}-\frac{2d_{1}}{t}+v_{1}}{t}=\frac{2v_{1}t-2d_{1}}{t^{2}}$

$=\frac{2(v_{1}t-d_{1})}{t^{2}}.$

Substituting values we get:

$v_{0}=\frac{2d_{1}}{t}-v_{1}=\frac{2\cdot 40.0}{8.5}-2.8\approx 6.1\ m/s,$

$a=\frac{2(v_{1}-d_{1}t)}{t^{2}}=\frac{2\cdot(2.8\cdot 8.5-40.0)}{8.5^{2}}\approx-0.45\ m/s^{2}.$

Answer. $v_{0}=6.1\ m/s,\quad a=-0.45\ m/s^{2}.$