Question

Aditya started driving his car . He increases the speed till 4 seconds and then he kept
his car constant speed for 6 seconds. Then he decreased the speed of the car up to
another 6 seconds . After reaching at the starting place , he draws the speed-time graph
of his 16 seconds driving as shown below:

) What type of motion is represented by OA ?
(a) Uniform velocity (b) Uniform acceleration (c) Negative acceleration (d) No acceleration
(ii) What type of motion is represented by BC ?
(a) Uniform velocity (b) Uniform acceleration (c) Negative acceleration (d) No acceleration
(iii) Find the acceleration of the body from 0 to 4 seconds.
(a) 1.5 m/s2
(b) 2 m/s2
(c) 3 m/s2
(d) 1 m/s2
(iv) Calculate the retardation of the body .
(a) 1.5 m/s2
(b) 2 m/s2
(c) 3 m/s2
(d) 1 m/s2
(v) Find out the distance travelled by the body from A to B .
(a) 15 m (b) 30 m (c) 36 m (d) 60 m

Answer 1

Given: Aditya started driving his car . He increases the speed till 4 seconds and then he kept his car constant speed for 6 seconds. Then he decreased the speed of the car up to another 6 seconds . After reaching at the starting place , he draws the speed-time graph of his 16 seconds driving as shown.

To find:

(i) The type of motion is represented by OA.

(ii) The type of motion is represented by BC.

(iii) The acceleration of the body from 0 to 4 seconds.

(iv) The retardation of the body.

(v) The distance travelled by the body from A to B.

Solution:

(i)

The speed of the car in the first 4 seconds is increasing uniformly with respect to time. This means that the acceleration is uniform. So, the motion represented by OA is uniform acceleration.

(ii)

The speed of the car is decreasing with respect to time in BC. This means that the car is undergoing retardation. So, the type of motion in BC is negative acceleration.

(iii)

The acceleration of the car can be given as follows.

$a = \frac{s}{t}$

Here, a is the acceleration, S is the speed and t is the time.

$a = \frac{6ms^{-1}}{4s}$

  $= 1.5 ms^{-2}$

(iv)

The retardation of the body can be calculated as follows.

$retardation = \frac{6 ms^{-1}}{6s}$

                   $= 1 ms^{-2}$

(v)

The distance travelled by the card can be calculated using the following formula.

$d = s* t$

  $= 6 ms^{-1} * 6s$

  $= 36 m$

Therefore,

(i) The type of motion is represented by OA is Uniform acceleration.

(ii) The type of motion is represented by BC is Negative acceleration.

(iii) The acceleration of the body from 0 to 4 seconds is 1.5 ms⁻².

(iv) The retardation of the body is 1 ms⁻².

(v) The distance travelled by the body from A to B is 36 m.

Although a figure of your question is missing, you might be referring to the one attached.

Answer 2

Answer:

Step-by-step explanation:Aditya started driving his car He mereuses the speed till 4 seconds and then he kept his card in constant speed for 6 seconds. Then after he decreased the speed of the car upto another seconds Ader reaching at the starting place, he draws the speed-time graph of his 16 seconds driving as shown below

1) What type of motion is represented by OA7

i) uniform velocity

ii) uniform acceleration

ii) negative acceleration

iv) no acceleration: