Question

Ok so I have a question which I don't quite understand. It's kinematics one.

A train is coming at the station. It starts deccelerating and passes first 50m in 5 seconds, and next 50 m in 7 s. Find acceleration and the starting velocity. I've studied it through, but cannot find the way to answer. I've got all formulas and equations.

Answer 1

Answer on Question#37994 - Physics - Mechanics

A train is coming at the station. It starts decelerating and passes first 50m in 5 seconds, and next 50 m in 7 s. Find acceleration and the starting velocity.

Solution:

a – deceleration of the train;

$V_0$ – initial velocity of the train;

Equation of motion for the train for first distance $d_1 = 50m$ ($t_1 = 5s$):

$$d_1 = V_0 t_1 - \frac{a t_1^2}{2}$$ $$V_0 = \frac{2 d_1 + a t_1^2}{2 t_1}$$

Rate equation for the train for this distance ($V_1$ – velocity after time $t_1$):

$$V_1 = V_0 - a t_1$$

Equation of motion for the train for first distance $d_2 = 50m$ ($t_2 = 7s$):

$$d_2 = V_1 t_2 - \frac{a t_2^2}{2}$$

(3) and (2) in (4):

$$d_2 = (V_0 - a t_1) t_2 - \frac{a t_2^2}{2} = \left(\frac{2 d_1 + a t_1^2}{2 t_1} - a t_1\right) t_2 - \frac{a t_2^2}{2}$$ $$2 d_2 t_1 = 2 d_1 t_2 + a t_1^2 t_2 - 2 a t_1^2 t_2 - a t_2^2 t_1$$ $$2 d_2 t_1 = 2 d_1 t_2 - a t_1^2 t_2 - a t_2^2 t_1$$ $$a = \frac{2 (d_1 t_2 - d_2 t_1)}{t_1 t_2 (t_1 + t_2)} = \frac{2 (50m \cdot 7s - 50m \cdot 5s)}{5s \cdot 7s \cdot (5s + 7s)} = 0.48 \frac{m}{s^2}$$ $$(2): V_0 = \frac{2 d_1 + a t_1^2}{2 t_1} = \frac{2 \cdot 50m + 0.48 \frac{m}{s^2} \cdot (5s)^2}{2 \cdot 5s} = 11.2 \frac{m}{s}$$

Answer: deceleration of the train is equal to $0.48 \frac{m}{s^2}$; starting velocity of the train is $11.2 \frac{m}{s}$.