Question

a train accelerates from 10km/hour to 40km/hour in 2 minutes. How much distance does it cover in this period? Assume that the tracks are straight

Answer 1

Answer:

Distance s=829.586 m

Explanation:

The initial velocity of the train is u=10 km/hr=$\dfrac{10\times 1000}{60\times 60} =2.77\ m/s$

The final velocity of the train is v=40 km/hr=$\dfrac{40\times 1000}{60\times 60}=11.11\ m/s$

The time taken by the train is t=2 min=120 sec

Let the distance traveled by the train during acceleration be s

The acceleration of the train is,

$a=\dfrac{v-u}{t}\\a=\dfrac{11.11-2.77}{120}\\a=0.07\ m/s^2$

From the equation of motion

$v^2-u^2=2as\\\\11.11^2-2.7^2=2\times 0.007\times s\\\\s=\dfrac{11.11^2-2.7^2}{2\times 0.007}\\\\s=829.586\ m$

Answer 2

Answer:

829.586

Explanation:

this is the answer