Question

A car moving with constant acceleration covered the distance between two points 56.2 m apart in 5.32 s. Its speed as it passes the second point was 14.8 m/s. (a) What was the speed at the first point? (b) What was the acceleration? (c) At what prior distance from the first point was the car at rest?

Answer 1

Answer on Question #51379, Physics, Mechanics | Kinematics | Dynamics

The law of motion of the car is $x(t) = v_0 t + \frac{a t^2}{2}$ , $v = v_0 + a t$ , where $v_0$ is velocity at first point. Let $T = 5.32 \, \text{s}$ , $s = 56.2 \, \text{m}$ , $v_2 = 14.8 \, \frac{\text{m}}{\text{s}}$ . Hence, $s = v_0 T + \frac{a T^2}{2}$ and $v_2 = v_0 + a T$ .

a) One has a linear system of equations for $v_0, a$ . Substituting $a = \frac{v_2 - v_0}{T}$ from the second equation into the first equation, obtain $s = v_0 T + \frac{v_2 - v_0}{2T} T^2 = v_0 T + \frac{v_2 - v_0}{2} T = \frac{1}{2} (v_2 + v_0) T$ , from where $\frac{2s}{T} = v_2 + v_0$ , thus $v_0 = \frac{2s}{T} - v_2 = 6.38 \frac{\text{m}}{\text{s}}$ - that is the speed of the car at first point.

b) Using $a = \frac{v_2 - v_0}{T}$ , obtain $a = \frac{14.8\frac{m}{s} - 6.38\frac{m}{s}}{5.32s} \approx 1.58\frac{m}{s^2}$ .

c) $v(t') = v_0 + at'$ , hence if $v(t') = 0 = v_0 + at'$ , from where $t' = \frac{-v_0}{a}$ (the time seems to be negative because we chose $t = 0$ at first point). Hence, the car had zero speed when it had coordinate $x(t') = \frac{-v_0^2}{a} + \frac{v_0^2}{2a} = \frac{-v_0^2}{2a} = -12.88m$ . The distance from the first point is thus $12.88m$ .

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