Question

speed a train increase from 36km/h to 72km/h in 10 minutes find distance covered by the train during this time​

Answer 1

9 km

Step-by-step explanation:

Given:

Initial velocity of the train = 36 km/h

Final velocity of the train = 72 km/h

time = 10 min = 10 *1/60 h = 1/6 h

To find:

Distance covered during this time.

Let us solve by using Equations of motion.

By first eqn of motion,

• v = u + at("+" because the train is accelerating)

Let us put values to find acceleration.

• 72 = 36 + a*1/6
• 72 -36 = a*1/6
• 36 = a/6
• a = 216 km/h²

To find distance , use second eqn of motion:

• S(Distance) = ut + 1/2 a*t²
• S = 36 *1/6 +1/2 *216 *1/6 *1/6
• S = 6 + 1/2* 6³/6²
• S = 6 + 1/2* 6 = 6 + 3 = 9 km

Abbreviations:

a = acceleration

v = final velocity

u = initial velocity

t = time taken during change in velocity

S = distance covered during this time

Answer 2

`8\; {\rm km}`, assuming that the train accelerated at a constant rate.

Step-by-step explanation:

Assume that the train accelerated at a constant rate, meaning that acceleration is constant. Displacement
`x` of the train can be found using the SUVAT equation:

`\displaystyle x = \left((u + v)/(2)\right) \, t`,

Where:

• 
  `u` is the velocity of the train before accelerating,
• 
  `v` is the velocity of the train after accelerating, and
• 
  `t` is the duration of the acceleration.

Make sure the unit of time is consistent. Since velocity is measured in kilometers per hour, duration of the acceleration should be measured in hours (not in minutes.)

`\begin{aligned} t &= 10\; \text{minute} \\ &= 10\; \text{minute} * \frac{1\; \text{hour}}{60\; \text{minute}} \\ &= (1)/(6)\; \text{hour}\end{aligned}`.

Given that
`u = 36\; {\rm km\cdot h^(-1)}` and
`v = 72\; {\rm km \cdot h^(-1)}`, the displacement of the train would be:

`\begin{aligned}x &= \left(\frac{36\; {\rm km\cdot h^(-1)} + 72\; {\rm km\cdot h^(-1)}}{2}\right)\, \left((1)/(6)\; \text{hour}\right) \\ &= 8\; {\rm km}\end{aligned}`.

In other words, the train would travel
`8\; {\rm km}` over the given period of time.