Question

A train slows down from 40ms-1 to 10ms-1 in 60 seconds, how far does it travel in that time?

Answer 1

Answer:

Here, Side of the given square field = 10m

so, perimeter of a square = 4*side = 10 m * 4 = 40 m

Farmer takes 40 s to move along the boundary.

Displacement after 2 minutes 20 s = 2 * 60 s + 20 s = 140 seconds

since in 40 s farmer moves 40 m

Therefore, in 1s the distance covered by farmer = 40 / 40 m = 1m

Therefore, in 140s distance covered by farmer = 1 � 140 m = 140 m. Now, number of rotation to cover 140 along the boundary= Total Distance / Perimeter

= 140 m / 40 m = 3.5 round

Thus, after 3.5 round farmer will at point C of the field.

Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from initial position.

Answer 2

The distance travelled by the train, s = 1.5 kms (1500 m).

Given,

Initial velocity of the train, u = 40 m/s

Final velocity of the train, v = 10 m/s

Time taken by the train, t = 60 s

To Find,

Distance travelled by the train, s.

Solution,

We have been given that a train slows down from 40 m/s to 10 m/s in 60 seconds.

We have to find the distance travelled by the train while decelerating.

Deceleration is the negative of acceleration. It is the negative rate of change of the velocity of an object with respect to time.

Mathematically, acceleration is expressed as:

⇒ a = (v-u)/t

where u is the initial velocity, v is the final velocity, and t is the time taken.

In the given scenario, the negative acceleration is given by:

⇒ a = (10 - 40)/60

⇒ a = - 30/60

⇒ a = - 1/2 m/s²

Now, the displacement s is given by one of the equations of motion:

⇒ s = ut + 1/2 at²

⇒ s = (40 × 60) + (1/2 × -1/2 × 60²)

⇒ s = 2400 - 1/4 × 3600

⇒ s = 2400 - 900

⇒ s = 1500 m

⇒ s = 1.5 kms

Hence, the distance travelled by the train = 1.5 kms (1500 m).

Read more on Brainly.in:

1. https://brainly.in/question/6780241

2. https://brainly.in/question/17908194

#SPJ2