Question

This problem is best done graphically. A ''toy train'' on its way to Darjeeling is speeding up slowly from v=0 to v=40.0 km/hr in 30.0 min and then slows down to a stop over the remaining 20.0 min. Plot v vs t for the train. Using your plot, calculate the distance traveled by the train in 50.0 min. Report the answer accurate to three significant figures.

-----km

B) What is the acceleration of the train in the first 30 min. Express your answer accurate to one significant figure

C.)What is the acceleration in the next 20 min. Express your answer accurate to one significant figure.

---- ×10−3 m/s2

Answer 1

Answer on Question #85034 – Physics – Mechanics | Relativity

This problem is best done graphically. A "toy train" on its way to Darjeeling is speeding up slowly from $v = 0$ to $v = 40.0 \, \text{km/hr}$ in 30.0 min and then slows down to a stop over the remaining 20.0 min. Plot $v$ vs $t$ for the train. Using your plot, calculate the distance traveled by the train in 50.0 min. Report the answer accurate to three significant figures.

---km

B) What is the acceleration of the train in the first 30 min. Express your answer accurate to one significant figure

C.) What is the acceleration in the next 20 min. Express your answer accurate to one significant figure.

---×10-3 m/s2

Solution

First $t_1 = 30$ min train speeding up slowly with positive uniform acceleration which correspond to the AB line on figure 1. After that he slows down to a stop with negative uniform acceleration (BC line) over the remaining $t_2 = 20.0$ min.

A. To calculate the distance traveled by the train in 50.0 min we need to find a area of triangle $ABC$ . Its more convenient to do by separating it into the two different triangles $ABD$ and $BCD$ . So full area will be the sum of their areas.

Let us denote $S_{1}$ as area of the triangle ABD and $S_{2}$ as area of the triangle BDC, so area of the $ABC$ is $S = S_{1} + S_{2}$

Using formula's for right triangle $S = \frac{1}{2} ab$ where $a$ and $b$ are the legs of the triangle

We obtain $S = \frac{1}{2} V t_{1} + \frac{1}{2} V t_{2} = \frac{1}{2} V\left(t_{1} + t_{2}\right)$ , where $V = 40 \, \text{km/h}$ is the maximum train's speed.

$$S = \frac {1}{2} V \left(t _ {1} + t _ {2}\right) = \frac {1}{2} 4 0 [ k m / h ] \left(\frac {1}{2} [ h ] + \frac {1}{3} [ h ]\right) = \frac {5 0}{3} [ k m ] \approx 1 6. 7 [ k m ]$$

B. Acceleration of the train in the first 30 min correspond to the tangent of angle $\alpha$

$$a = \tan (\alpha) = \frac {V}{t _ {1}} = \frac {4 0 \cdot 1 0 0 0 / 3 6 0 0 [ m / s ]}{3 0 \cdot 6 0 [ s ]} = \frac {1}{1 6 2} [ m / s ^ {2} ] \approx 0. 0 0 6 [ m / s ^ {2} ]$$

where we use $[km / h] = \frac{1000}{3600} [m / s]$

C. Analogously, acceleration in the next 20 min correspond to the tangent of angle $\pi -\beta$

$$a = \tan (\pi - \beta) = - \tan (\beta) = - \frac {V}{t _ {2}} = \frac {4 0 \cdot 1 0 0 0 / 3 6 0 0 [ m / s ]}{2 0 \cdot 6 0 [ s ]} = - \frac {1}{1 0 8} [ m / s ^ {2} ] \approx - 0. 0 0 9 [ m / s ^ {2} ],$$

Answer: (A) = 16.7, (B) 6, (C) -9

https://en.wikipedia.org/wiki/Right_triangle