Question

study the speed time graph of a body shown in the figure below. Calculate
i)distance travelled from o to a
ii) distance travelled from b to c
iii)the total distance travelled by the body in 16 secs
iv)what is the distance travelled by the body when the body is moving with uniform speed.
v)state the type of motion of OA,AB and BC
vi) calculate a of OA
vii)calculate retardation

Answer 1

Answer:

i)12m

ii)36m

iii)(12+12+36)m =50m

iv)12m

vi)6/4 m/s^2

    =1.5 m/s^2

vii)-1.5 m/s^2

(FOR DETAIL ANSWER FOLLOW AND DM ME ON IG-karsiddharth7)

Answer 2

Answer:

Explanation:

We know that  

                          $\frac{ds}{dt}$=v

So the area of v-t graph will give the distance.

(1) The area of graph from 0 to a=12

(2) The area of graph from b to c=12

(3)The area of graph from 0 to c=12+36+12

                                                    =60

(4)Uniform velocity means that velocity is constant .And not

varying with time .

So easily we can see that from graph from a to b velocity is uniform

So distance travel between them =36

(4) OA is the motion when velocity is increasing with time it means that

acceleration.

AB is the motion when velocity is uniform it means that acceleration is zero

BC s the motion when velocity is decreasing with time it means that

motion is retardation.

(5) $a=1.5 m/s^2$