derive an expression for the distance travelled by a body moving with a uniform motion "a" having initial velocity "u" and final velocity "v" after a time "t" from its velocity - timegraph.
Answers
Explanation:
Let
′
u
′
be the initial velocity
′
v
′
be the velocity at time
′
t
′
and
′
a
′
be the acceleration.
The distance travelled by the body in
′
t
′
second = area enclosed between the velocity-time graph and x−axis
Or
Distance traveled by the body in
′
t
′
seconds = area of trapezium OABD
=1/2×(sum of parallel sides)×height
=1/2×(u+v)×t
=
2
(u+v)t
=
2
(u+at)t
=u+
2
1
at
2
Explanation:
Let
′
u
′
be the initial velocity
′
v
′
be the velocity at time
′
t
′
and
′
a
′
be the acceleration.
The distance travelled by the body in
′
t
′
second = area enclosed between the velocity-time graph and x−axis
Or
Distance traveled by the body in
′
t
′
seconds = area of trapezium OABD
=1/2×(sum of parallel sides)×height
=1/2×(u+v)×t
=
2
(u+v)t
=
2
(u+at)t
=u+
2
1
at
2