Study the speed time graph of a body shown in Figure. and answer the following questions: (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) Find out acceleration of the body. (e) Find out retardation of the body . (f) Find out the distance travelled by the body from `A` to `B`

Study the speed time graph of a body shown in Figure. and answer the f

1.	Study the speed time graph of a body shown in Figure. and answer the following questions: (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) Find out acceleration of the body. (e) Find out retardation of the body . (f) Find out the distance travelled by the body from `A` to `B`
Answer» Correct Answer - `(a) uniformly accelerated , (b) uniform velocity , ( c ) uniformly retarded , (d) `1.5 m//s^(2)` , (e) `36 m` (a) As speed - time graph `OA` is a straight line with a positive slope , the motion is uniformly accelerated motion. (b) As `AB` is parallel to time axis , its slope is zero . Therefore , acceleration of the body is zero . The body is moving with a uniform velocity. ( c ) As speed - time graph `BC` is a straight line with negative slope, the motion is uniformly decelerated motion , (d) From `O` to `A` , change in speed `= 6 - 0 = 6 m//s` time taken `= 4 - 0 = 4 s` Thus , acceleration `= ("change in speed")/("time taken") = (6 m//s) /( 4 s) = 1.5 m//s^(2)` (e) From `B` to `C` change in speed `= 0 - 6 = - 6 m//s` time taken `= 16 - 10 = 6 s` Thus , acceleration `= ("change in speed")/("time taken") = ( - 6 m//s)/( 6 s) = - 1 m//s^(2)` Negative sign indicates retardation. (f) From `A` to `B` , the velocity is uniform. `:.` distance travelled ` = "velocity" xx "time" = 6 xx ( 10 - 4) = 36 m`