- Equation For Velocity-Time Relation:
Consider the velocity-time graph of an object that moves under uniform acceleration. Initial velocity of the object is u and then it increases to v in time t. The velocity changes at a uniform rate a.
The perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time interval t.
Let us draw AD parallel to OC. From the graph, we observe that BC = BD + DC = BD + OA
Substituting BC = v and OA = u,
we get v = BD + u i.
or BD = v – u ii.
From the velocity-time graph,
The acceleration of the object is given by a = Change in velocity/time taken
= BD/AD = BD/OC
Substituting OC = t, we get
a = BD/t
BD = at iii.
Using the equations ii. & iii. We get,
v = u + at
- Equation For Position - Time Relation:
Let us consider that the object has travelled a distance s in time t under uniform acceleration a. The distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB. Thus, the distance s travelled by the object is given by
s = area OABC (which is a trapezium)
= area of the rectangle OADC + area of the triangle ABDM.
= OA × OC + 1/2 (AD × BD)
Substituting OA = u, OC = AD = t and BD = at, we get
s = u × t + 1/2 (t×at)
or
s = u t + 1/2 a t2
- Equation For Position - Velocity Relation:
From the velocity-time graph, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph. That is,
s = area of the trapezium OABC = (OA + BC) X OC/2
Substituting OA = u, BC = v and OC = t,
We get
s= (u+v) t/2
From the velocity-time relation
We get
t= v - u/a
We have
s = (v + u) x (v – u)/2a
or 2as = v2 – u2
Meghna Thapar 3 years, 11 months ago
Consider the velocity-time graph of an object that moves under uniform acceleration. Initial velocity of the object is u and then it increases to v in time t. The velocity changes at a uniform rate a.
The perpendicular lines BC and BE are drawn from point B on the time and the velocity axes respectively, so that the initial velocity is represented by OA, the final velocity is represented by BC and the time interval t is represented by OC. BD = BC – CD, represents the change in velocity in time interval t.
Let us draw AD parallel to OC. From the graph, we observe that BC = BD + DC = BD + OA
Substituting BC = v and OA = u,
we get v = BD + u i.
or BD = v – u ii.
From the velocity-time graph,
The acceleration of the object is given by a = Change in velocity/time taken
= BD/AD = BD/OC
Substituting OC = t, we get
a = BD/t
BD = at iii.
Using the equations ii. & iii. We get,
v = u + at
Let us consider that the object has travelled a distance s in time t under uniform acceleration a. The distance travelled by the object is obtained by the area enclosed within OABC under the velocity-time graph AB. Thus, the distance s travelled by the object is given by
s = area OABC (which is a trapezium)
= area of the rectangle OADC + area of the triangle ABDM.
= OA × OC + 1/2 (AD × BD)
Substituting OA = u, OC = AD = t and BD = at, we get
s = u × t + 1/2 (t×at)
or
s = u t + 1/2 a t2
From the velocity-time graph, the distance s travelled by the object in time t, moving under uniform acceleration a is given by the area enclosed within the trapezium OABC under the graph. That is,
s = area of the trapezium OABC = (OA + BC) X OC/2
Substituting OA = u, BC = v and OC = t,
We get
s= (u+v) t/2
From the velocity-time relation
We get
t= v - u/a
We have
s = (v + u) x (v – u)/2a
or 2as = v2 – u2
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