Answer to Question #132107 in Molecular Physics | Thermodynamics for Abdulazeez

Question

A moving train travels through station P with a speed of 20 meter per seconds anf accelerate uniformly for 2 seconds reaching a speed of 48 meter per second, it continue with this speed for 5 seconds and then retardate uniformly for another 3 seconds and comes to rest at station Q. Find the distance from P to Q

Answer 1

Solution:

acceleration :

a = $\tfrac{(V_{f} - V_{i})}{2}$ = $\tfrac{48-20}{2} =14 (\tfrac{m}{s^{2}} )$

acceleration time :

t_a = 2.0 sec

retardation :

r = $\tfrac{0-48}{3}=-16 (\tfrac{m}{s^{2}} )$ = -16 m/sec^2

retardation time:

t_r = 3.0 sec

distance computation :

$S_{1}=V_{i}\cdot t_{a}\cdot \tfrac{a\cdot t_{a}^{2}}{2} =68m$

$S_{2}=V_{f}\cdot 5=240m$

$S_{3}=V_{f}\cdot \tfrac{t_{r}}{2} =72m$

$S_{total}=S_{1}+S_{2}+S_{3}=380m$

The distance from P to Q is 380 m.