40 km/h = 0.278 m/s
If it accelerated at a m/s^2, then it took t=0.278/a seconds to achieve that speed.
The distance covered while accelerating/decelerating is thus
1/2 at^2 + 0.278(3t) - 1/2 (a/3)(3t)^2
= 1/2 a(0.278/a)^2 + 0.278(3(0.278/a))-1/2 (a/3)(3(0.278/a))^2
= 0.155/a
Now, if it travels at constant speed for x seconds, we have
0.155/a + 0.278x = 3000
4(0.278/a) + x = 300
Solve for a. As always, double-check my math.
a train takes 5 minutes to cover a distance of 3 km between two stations p and q . Starting from rest at p , it accelerate at a constant rate to a speed 40 km/h and maintain this speed untill it is brought to rest at q . If train takes 3 times as long to deaccelerate as it does to accelerate , find time taken by the train to accelerate .
2 answers
Big OOOPS! I used a speed of 1 km/hr, not 40. See Damon's "physics" solution below.
2 answers