Question

A train accelerated from 10km/hr to 40km/hr in 2 minutes. How much distance does it cover in

this period? Assume that the tracks are straight?​

Answer 1

Given :

▪ Initial velocity of train = 10kmph

▪ Final velocity of train = 40kmph

▪ Time interval = 2min

To Find :

▪ Distance covered by the train in the given interval of time.

Solution :

→ Since, acceleration has said to be constant throughout the journey, we can easily apply equation of kinematics.

→ First we have to find out acceleration of train after that we can calculate distance covered by train.

____________________________

• 1kmph = 5/18mps
• 10kmph = 10×5/18 = 2.78mps
• 40kmph = 11.11mps
• 2min = 2×60 = 120s

☞ v = u + at

☞ 11.11 = 2.78 + 120a

☞ 120a = 8.33

☞ a = 0.07m/s^2

✒ S = ut + (1/2)at^2

✒ S = (2.78×120)+[0.5×0.07×(120)^2]

✒ S = 333.6 + 504

✒ S ≈ 838m

Answer 2

Answer:

$\frac{5}{72} \ m/s^{2}$, 838 metres

Explanation:

Given:

Initial velocity of the train = u = 10 km/hr

10 km/hr = $10 \times \frac{5}{18} = \frac{25}{9} m/s$

Final velocity of the train = v = 40 km/hr

40 km/hr = $40 \times \frac{5}{18} = \frac{100}{9} m/s$

Time = 2 minutes

1 minute = 60 seconds

2 minutes = 2×60 seconds = 120 seconds

To find:

Acceleration (a)

Acceleration = $\frac{v-u}{t} ; where \ v=Final\ velocity , u=initial \ velocity\ and \ t=time$

Substituting the values, we get:

Acceleration = $\frac{\frac{100}{9}-\frac{25}{9} }{120}$

Acceleration = $\frac{\frac{75}{9} }{120}$

Acceleration = $\frac{75}{9} \times \frac{1}{120}$

Acceleration = $\frac{25}{360}$

Acceleration = $\frac{5}{72} \ m/s^{2}$

The acceleration of the train is equal to $\frac{5}{72} \ m/s^{2}$

Distance can be found third equation of motion which says:

v²-u²=2as

$\frac{10000}{81} -\frac{625}{81} = 2 \times \frac{5}{72} \times s$

$\frac{9375}{81} = \frac{10}{72} s$

s= $\frac{9375}{81} \times \frac{72}{10}$

s= 838 metres