Question

study the speed time graph of a body shown in above figure and answer the following:-.
(a)What type of motion is represented by OA?
(b)What type of motion is represented by AB?
(c)What type of motion is represented by BC?
(d) calculate the acceleration of the body.
(e) calculate the retardation of the body.
(f) calculate the distance travelled by body from A to B​

Answer 1

Answer:

a) The type of motion represented by OA is :

Uniformly Accelerated Motion

b) The type of motion represented by AB is :

Uniform motion with constant speed

c) The type of motion represented by BC is :

Uniformly Retarded Motion.

d) Acceleration of the body is represented in the stretch between O and A . Acceleration is defined as the change in velocity with respective change in time.

$acc. = \dfrac{6 - 0 }{4 - 0} = 1.5 \: m {s}^{ - 2}$

e) Retardation of the body is represented in the stretch between B and C . Retardation is also a type of Acceleration but in the opposite direction wrt velocity.

$retd. = \dfrac{0 - 6 }{16 - 10} = - 1\: m {s}^{ - 2}$

f) Distance travelled between A and B will be equal to the area under the A-B part of the graph.

$distance = area \: of \: rectangle$

$= > distance = 6 \times (10 - 4)$

$= > distance = 36 \: m$

Answer 2

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QUESTION :

Study the speed time graph of a body shown in above figure and answer the following:-.

(a)What type of motion is represented by OA?

(b)What type of motion is represented by AB?

(c)What type of motion is represented by BC?

(d) calculate the acceleration of the body.

(e) calculate the retardation of the body.

(f) calculate the distance travelled by body from A to B.

SOLUTION :

1. At OA -

Here, the graph shows uniform accelerated motion.

Here,

Acceleration = {6 - 0 } / 4 = 1.5 m / s ^ 2

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2. At OA, the object is at rest.

There is no acceleration or deceleration.

3. At BC, the object is showing uniform deceleration.

Deceleration = 6 / 6 = 1 m / s ^ 2

4. The acceleration of the body is 1.5 m / s ^ 2 as calculated before.

5. The magnitude of retardation of the body is is -1 m / s ^ 2.

6. The distance travelled by the body between AB can be calculated by calculating the area as this is a speed time graph.

Distance = 6 × 6 = 36 m.

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