Answer to Question #165666 in Physics for Bhavya

Question

A body covers 10 m at 4 m/s and covers rest of the journey at uniform speed. If the average speed of the body to cover the complete journey of 15 m is 5m/s², calculate its speed after covering 10 m.

Answer 1

I assume that there is a mistake in units and it should be $a = 4 m/s^2$ and $v_{av} = 5 m/s$ . Then, by definition of average speed, the time required to complete the journey is:

$$t_{total} = \dfrac{15m}{5m/s} = 3s$$

Let's denote the speed after covering $d_1 = 10m$ as $v_2$. Then, by definition of the speed of motion with uniform acceleration, obtain:

$$v_2 = v_1 + at_1$$

where $v_1$ is the speed at the very beginning of the motion, and $t_1$ is the time required to cover $d_1$.

The distance covered under the uniform acceleration is given as follows:

$$d_1 = v_1t_1 + \dfrac{at_1^2}{2}$$

Substituting here $v_1$ from the previous equation, obtain:

$$d_1 = (v_2 - at_1)t_1 + \dfrac{at_1^2}{2} = v_2t_1 - \dfrac{at_1^2}{2}$$

Since the second part of the journey $d_2 = 5m$ was covered with uniform spee, then have:

$$v_2 = \dfrac{d_2}{t_2} = \dfrac{d_2}{t_{total} - t_1}$$

where $t_2 = t_{total} - t_1$ is the time required to cover $d_2$ with uniform speed $v_2$. Expressing $t_1$ from the last equation and substituting it into the previous one, obtain:

$$t_1 = t_{total} - \dfrac{d_2}{v_2}\\ d_1 = v_2\left( t_{total} - \dfrac{d_2}{v_2} \right) - \dfrac{a}{2}\left( t_{total} - \dfrac{d_2}{v_2} \right)^2=\\ =v_2t_{total} - d_2 - \dfrac{a}{2}\left( t_{total}^2 - 2\dfrac{t_{total}d_2}{v_2} + \dfrac{d_2^2}{v_2^2} \right)$$

Multiplying by $v_2^2$, obtain:

$$v_2^3t_{total} - v_2^2\left( d_1 + d_2 + \dfrac{at_{total}^2}{2} \right) - v_2at_{total}d_2 + \dfrac{ad_2^2}{2} = 0\\ v_2^3\cdot 3 - v_2^2\left( 10 +5 + \dfrac{4\cdot 3^2}{2} \right) - v_2\cdot 4\cdot 3\cdot 5+ \dfrac{4\cdot 5^2}{2} = 0\\ 3v_2^3 -33 v_2^2 - 60v_2+ 50 = 0$$

Solving this equation, obtain:

$$v_2 \approx 12.49\space \text{or} \space v_2 \approx 0.63\space \text{or} \space v_2 \approx-2.12$$

The speed was assumed to be positive, thus, we can ignore the last value. The second value does not fit as well, since then it is required to spend $t_2 = d_2/v_2 \approx 8s$ to cover the second part, which is more than total time of journey $t_{total}$. Thus, the only value remain is

$$v_2 \approx 12.49m/s$$

Answer. 12.49 m/s.