Question

Consider a mass initially moving at 7.50 m/s. How does it take to move 3.5 km (Be sure to convert to meters) if it accelerates at .550 m/s^2.?
Please show your work.

Answer 1

The mass is moving by uniformly accelerated motion, with initial velocity $v_i=7.50 m/s$ and acceleration $a=0.55 m/s^2$. Its position at time t is given by the following law:
$x(t)=v_i t + \frac{1}{2}at^2$
where we take the initial position $x_i=0$ since we are only interested in the distance traveled by the mass.

If we put $x(t)=d=3.5 km=3500 m$ into the equation, the corresponding time t is the time it takes for the mass to travel this distance:
$\frac{1}{2}at^2+v_it-d=0$
$4.9t^2+7.5t-3500 =0$
And the two solutions for the equation are:
$t=-25.5 s$ --> negative, we can discard it
$t=24.0 s$ --> this is the solution to our problem