Question

A bus traveling at 40 m/s accelerates at a constant rate of 3.5 m/s2 over a distance of 50 m. What is its speed after accelerating?

Answer 1

We know that

• The initial speed is 40 m/s.

,

• The acceleration is 3.5 m/s^2.

,

• The distance covered is 50 m.

Let's use the following formula to find the final speed.

$v^2_f=v^2_0+2ad$

Where V0 = 40 m/s, a = 3.5 m/s^2, and d = 50m.

$\begin{gathered} v^2_f=(40\cdot\frac{m}{s})^2+2(3.5\cdot\frac{m}{s^2})(50m) \\ v^2_f=1600\cdot\frac{m^2}{s^2}+350\cdot\frac{m^2}{s^2} \\ v^2_f=1950\cdot\frac{m^2}{s^2} \\ v_f=\sqrt[]{1950\cdot\frac{m^2}{s^2}} \\ v_f\approx44\cdot\frac{m}{s} \end{gathered}$

Therefore, the speed after accelerating is 44 m/s.