study the speed time graph of a body given here 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 the acceleration of the body.
e) calculate the retardation of the body.
f) find out the distance travelled by the body from A and B.
Answers
Given:
Time loops = 4, 10 and 16
To Find:
Motions represented in OA, AB and BC and the distance travelled from A to B.
Solution:
1. In the region OA - The velocity increases uniformly with time, thus it represents uniform acceleration.
2. In the region AB - The body moves with a constant speed, and hence moving with zero acceleration.
3. In the region BC - The velocity decreases uniformly with time, and represents a uniform retardation.
4. Acceleration of the body OA = 6-0/ 4-0
= 6/4 = 1.5m/s²
5. Retardation in the body = 6-0/16-0
= 6/16 = 1m/s²
By observing the given speed-time graph, we can observe the following:
(a) OA represents motion with uniform acceleration
OA represents the motion of the body where its velocity increases from 0 m/s to 6 m/s, from time 0 s to 4 s. Also, speed and time are directly proportional here. Hence, the OA represents a positive acceleration of the body, and since it is a straight line, the acceleration here is uniform.
(b) AB represents motion with uniform speed
The motion represented by AB does not result in the change of velocity, instead, we see a straight line parallel to the x-axis. Therefore, the speed remains 6 m/s from 4 s to 10 s. Hence, the motion is uniform, and it is not accelerated.
(c) BC represents motion with uniform negative acceleration (uniform retardation)
BC represents the motion of the body where its velocity decreases from 6 m/s to 0 m/s, from time 10 s to 16 s. Also, speed and time are inversely proportional here. Hence, the BC represents a negative acceleration (or retardation) of the body, and since it is a straight line, the negative acceleration here is uniform.
(d) The acceleration of the body is 1.5 m/s²
The acceleration of a body is given by the slope of the velocity-time graph. Here, the slope is given by:
⇒ a = AD ÷ OD
⇒ a = (6-0) m/s ÷ (4-0) s
⇒ a = 6 m/s ÷ 4 s
⇒ a = 6 m/s²
⇒ a = 3/2 m/s²
⇒ a = 1.5 m/s²
(e) The retardation of the body is 1 m/s²
The acceleration of a body is given by the slope of the speed-time graph. Retardation is the negative of acceleration. Here, the slope is given by:
⇒ a = BE ÷ BC
⇒ a = (0-6) m/s ÷ (16-10) s
⇒ a = -6 m/s ÷ 6 s
⇒ a = -1 m/s²
⇒ r = 1 m/s²
(f) The distance traveled by the body from A to B is 66 m.
The distance traveled by the body is given by the area of the speed-time graph.
Here, the area of OABCO = area of OAD + area of ABED + area of BEC
[area of OAD = 1/2 × OD × AD = 1/2 × 4 × 6 = 12]
[area of ABED = AD × AB = 6 × 6 = 36]
[area of BEC = 1/2 × BE × EC = 1/2 × 6 × 6 = 18]
⇒ d = 12 + 36 + 18
⇒ d = 66 m
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