Question

A train accelerated from 36km/he to 90km/hr in 4 minutes . How much distance does it cover in this period? Assume that the tracks are straight

Answer 1

Answer:

4.2 kilometres

Explanation:

Given :

• Initial velocity = u = 36 km/hr

• Final velocity = v = 90 km/hr

• Time taken = 4 minutes = 240 seconds

To find :

• Distance travelled in this time period

Initial velocity = 36×5/18 = 10 m/s

Final velocity = 90×5/18 = 25 m/s

Acceleration =(25-10)/240

Acceleration = 15/240

Acceleration = 0.0625 m/s²

Using the second equation of motion :

S=ut+½×at²

S=10×240+½×0.0625×240²

S=2400+1800

S=4200 metres

The distance covered by the train is equal to 4200 metres or 4. 2 kilometres

Answer 2

Given :

▪ Initial velocity of train = 36kmph

▪ Final velocity of train = 90kmph

▪ Time of journey = 4min

To Find :

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

Concept :

☞ Since, acceleration has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this question.

✴ First equation of kinematics :

$\bigstar\:\underline{\boxed{\bf{\red{v=u+at}}}}$

✴ Third equation of kinematics :

$\bigstar\:\underline{\boxed{\bf{\blue{v^2-u^2=2as}}}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance
• t denotes time

Conversion :

↗ 1kmph = 5/18mps

↗ 36kmph = 36×5/18 = 10mps

↗ 90kmph = 90×5/18 = 25mps

↗ 4min = 4×60 = 240s

Calculation :

$\dashrightarrow\sf\:v^2-u^2=2{\huge(}\dfrac{v-u}{t}{\huge)}\times s\\ \\ \dashrightarrow\sf\:(25)^2-(10)^2=2{\huge(}\dfrac{25-10}{240}{\huge)}\times s\\ \\ \dashrightarrow\sf\:625-100=\dfrac{30}{240}\times s\\ \\ \dashrightarrow\sf\:s=\dfrac{525\times 240}{30}\\ \\ \dashrightarrow\underline{\boxed{\bf{\green{s=4200m=4.2km}}}}\:\orange{\bigstar}$