Physics, asked by tamil123, 1 year ago

A car starts accelerating from rest to a velocity of 20 m/s in 10 min, maintains the velocity for 4 min and then comes to rest with uniform deceleration in 3 min. Draw the v–t graph. Also calculate the total distance travelled by the car.

Answers

Answered by danielochich
3
When we draw the graph we find that the motion is in form of a trapezium.

To calculate the distance we should understand that the area under the graph equals to distance.

This is because it is a graph of velocity against time and since distance equals to speed × time, we get the area.

In calculating the area we must understand that the time is in minutes and must be converted to seconds.

AREA :

We divide the trapezium into two triangles and one rectangle.

Areas of the figures are as follows :

Triangle one (during acceleration) :

0.5 × 20 × 600 = 6000 m

Triangle 2 (during declaration) :

0.5 × 20 × 180 = 1800 m

Rectangle (constant speed) :

20 × 40 = 800 m

Total distance :

6000 + 1800 + 800 = 8600 m

I have attached the diagram.
Attachments:

Pranothi1: Why did u multiply 40 with 20?
Answered by Anonymous
2

Answer:

Explanation:

When we draw the graph we find that the motion is in form of a trapezium.

To calculate the distance we should understand that the area under the graph equals to distance.

This is because it is a graph of velocity against time and since distance equals to speed × time, we get the area.

In calculating the area we must understand that the time is in minutes and must be converted to seconds.

AREA :

We divide the trapezium into two triangles and one rectangle.

Areas of the figures are as follows :

Triangle one (during acceleration) :

0.5 × 20 × 600 = 6000 m

Triangle 2 (during declaration) :

0.5 × 20 × 180 = 1800 m

Rectangle (constant speed) :

20 × 40 = 800 m

Total distance :

6000 + 1800 + 800 = 8600 m

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