Question

A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final speed of 2.80 m/s. (a) Find its original speed. (b) Find its acceleration.

Answer 1

B) Since $V_1=V_0-a\times t$ , where $V_1$ is the final velocity, $V_0$ is the initial velocity, then $V_0=V_1+a\times t$ ,

and since $S =\frac {V_1^2-V_0^2}{2\times a}$ ,

$S=\frac {V_1^2-V_1^2+2\times V_1\times a \times t + a^2 \times t^2}{2\times a}$

$a=\frac{2\times S- 2\times V_1 \times t }{t^2}=\frac{2\times 40- 2\times 2.8 \times 8.5}{8.5^2}\approx0.45$ m/s$^2$

а) $V_0=V_1+a\times t=2.8+0.45\times 8.5=6.625$ m/s