Derive the equations v = u + at, s = ut + ½at2 and v2 = u2 + 2as graphically.
2. What is uniform circular motion ? Give two examples which force is responsible for that.
3.What can you say about the motion of a body if its displacement-time graph and velocity-time graph both are straight line ?
Answers
Answer:
(1) derive : The equations ....
v = u + at
s = u + 1/2 at²
v² = u² + 2as
solution : let a particle moves with initial velocity u after time t, its velocity becomes v due to acceleration acting on particle is a.
see figure,
slope of velocity - time graph = acceleration
⇒(v - u)/(t - 0) = a
⇒v - u = at
⇒v = u + at .........(1)
area enclosed the velocity - time graph = displacement covered by particle
⇒area of trapezium formed as shown in figure = S
⇒S = 1/2 [v + u ] × t
from equation (1),
⇒S = 1/2 [u + at + u ] × t
⇒S = 1/2 [2u + at] × t
⇒S = ut + 1/2 at² ...........(2)
we know, acceleration, a = v dv/ds
⇒a ∫ds = ∫v dv
⇒a[s] = [v²/2]
⇒as = 1/2 [v² - u²]
⇒2as = v² - u²
⇒v² = u² + 2aS ............(3)
(2) The final motion characteristic for an object undergoing uniform circular motion is the net force. The net force acting upon such an object is directed towards the center of the circle. The net force is said to be an inward or centripetal force.
Explanation:
(3) displacement-time graph is a straight line inclined to the time axis at an acute angle, it means that gradient of the curve or velocity of the body is non-zero. Therefore, body is moving away from the starting point with uniform velocity.
Answer:
To derive : The equations ....
v = u + at
s = u + 1/2 at²
v² = u² + 2as
solution : let a particle moves with initial velocity u after time t, its velocity becomes v due to acceleration acting on particle is a.
see figure,
slope of velocity - time graph = acceleration
⇒(v - u)/(t - 0) = a
⇒v - u = at
⇒v = u + at .........(1)
area enclosed the velocity - time graph = displacement covered by particle
⇒area of trapezium formed as shown in figure = S
⇒S = 1/2 [v + u ] × t
from equation (1),
⇒S = 1/2 [u + at + u ] × t
⇒S = 1/2 [2u + at] × t
⇒S = ut + 1/2 at² ...........(2)
we know, acceleration, a = v dv/ds
⇒a ∫ds = ∫v dv
⇒a[s] = [v²/2]
⇒as = 1/2 [v² - u²]
⇒2as = v² - u²
⇒v² = u² + 2aS ............(3)
