Question

A train started from rest and moved with constant acceleration. At one time it was traveling 20 m/s, and 170 m farther on it was traveling 46 m/s. Calculate (a) the acceleration, (b) the time required to travel the 170 m mentioned, (c) the time required to attain the speed of 20 m/s, and (d) the distance moved from rest to the time the train had a speed of 20 m/s.

Answer 1

Answer on Question #51377, Physics, Mechanics | Kinematics | Dynamics

Question

A train started from rest and moved with constant acceleration. At one time it was traveling $20\mathrm{m/s}$, and $170\mathrm{m}$ farther on it was traveling $46\mathrm{m/s}$. Calculate (a) the acceleration, (b) the time required to travel the $170\mathrm{m}$ mentioned, (c) the time required to attain the speed of $20\mathrm{m/s}$, and (d) the distance moved from rest to the time the train had a speed of $20\mathrm{m/s}$.

Solution

(a) Since the train started from rest and moved with constant acceleration:

$$\begin{array}{l} V_{2} = V_{1} + at \\ V_{1} = 20\,m/s, \quad V_{2} = 46\,m/s \\ \end{array}$$

By the same time:

$$S = V_{1}t + \frac{at^{2}}{2}, \quad S = 170\,m$$

So, we have a system with 2 variables $a$ and $t$:

$$\begin{cases} S = V_{1}t + \frac{at^{2}}{2} \\ V_{2} = V_{1} + at \\ \end{cases}$$ $$\begin{aligned} t &= \frac{V_{2} - V_{1}}{a} \Rightarrow S = V_{1} \frac{V_{2} - V_{1}}{a} + \frac{a}{2} \left(\frac{V_{2} - V_{1}}{a}\right)^{2} \\ S &= \frac{V_{1} \left(V_{2} - V_{1}\right)}{a} + \frac{\left(V_{2} - V_{1}\right)^{2}}{2a} = \frac{2V_{1} \left(V_{2} - V_{1}\right) + \left(V_{2} - V_{1}\right)^{2}}{2a} \Rightarrow \\ a &= \frac{2V_{1} \left(V_{2} - V_{1}\right) + \left(V_{2} - V_{1}\right)^{2}}{2S} = \frac{40(46 - 20) + 26^{2}}{340} \frac{m^{2}}{s^{2}m} = \\ &= 5.047 \frac{m}{s^{2}} \\ \end{aligned}$$

(b) $t = \frac{V_{2} - V_{1}}{a} = \frac{46\,m/s - 20\,m/s}{5.047\,m/s^{2}} = 5.152\,s$

(c) Since the train speeds up at a constant rate from rest:

$$\begin{array}{l} V_{0} = 0 \\ V_{1} = V_{0} + a t_{0} = a t_{0} \\ t_{0} = \frac{V_{1}}{a} = \frac{20 \, \text{m/s}}{5.047 \, \frac{\text{m}}{\text{s}^{2}}} = 3,963 \, \text{s} \\ \end{array}$$

(d) $S_{0} = \frac{a t_{0}^{2}}{2} = 0.5 \cdot 5.047 \, \frac{\text{m}}{\text{s}^{2}} \cdot (3,963)^{2} \, \text{s}^{2} = 39,627 \, \text{m}$

**Answer**

(a) $5.047 \, \frac{\text{m}}{\text{s}^{2}}$

(b) $5.152 \, \text{s}$

(c) 3,963 s

(d) $39,627 \, \text{m}$

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