flipperworks.com Examples of displacement-time graphsK v 2. highest point si (c) s is max, v = 0n s vem 3. body falling. 1. As bodya goes up, s ↑ s ↓ but v ↑ int but v ↓. magnitudeic ts motion of a body thrown vertically up and then returning to the point of projection
flipperworks.com Velocity and Acceleration Definition GraphicallyVelocity Rate of ds Gradient of s-t change of v= graph displacement dtAcceleration Rate of Gradient of v-t dv change of velocity a= graph dt
flipperworks.comK [b] Velocity-time graphin shows the velocity of a body at anye instant of time.ti the gradient of the graph is thec instantaneous acceleration of the body.s the area under the graph is the displacement of the body.
flipperworks.com v B C A Area = A1 O E H t Area = A2 D F G v (+ve when pointing to the right)O
flipperworks.comv B C A Area = A1O E H t Area = A2 D F G O to A v increases from zero at constant rate ⇒ body is moving from rest with uniform acceleration
flipperworks.com v B C A Area = A1O E H t Area = A2 D F GA to Bv increasing but gradient decreasing⇒ body continues to move faster butwith decreasing acceleration.
flipperworks.com v B C A Area = A1 O E H t Area = A2 D F GB to Cv remains constant⇒ zero acceleration
flipperworks.com vK B Ci An Area = A1e E H O tm D Area = A2c F Gs C to D v decreases at a constant rate but still positive ⇒ acceleration is constant but negative i.e. constant deceleration.
flipperworks.com v B C A Area = A1 O E H t Area = A2 D F GD to Ev = 0 ⇒ body is stationary
flipperworks.com v Q: At time F, is B C displacement of car A negative? Area = A1 O E H t Area = A2 D F GE to Fv negative but magnitude is increasing atconstant rate ⇒ body is moving in oppositedirection and speeds up ⇒ uniform negativeacceleration.
flipperworks.com v Q: At time F, is B C displacement of car A negative? Area = A1 O E H t Area = A2 D F G v (+ve when pointing to the right)O A B C D HG F E
flipperworks.com vK B Ci An Area = A1e E H O ta D Area = A2t F Gic F to Gs v remains constant
flipperworks.comv B C A Area = A1O E H t Area = A2 D F G G to H v is negative but acceleration is positive ⇒ constant deceleration. Body slows down and comes to rest at H
flipperworks.com v B C A Area = A1 O E H t Area = A2 D F GTotal distance moved = A1 + A2Net displacement = A1 - A2
flipperworks.com Examples of velocity-time graphs(a) v (b) v 2 3 1 t t line 1: uniform velocity increasing acceleration line 2: uniform acceleration line 3: uniform deceleration
flipperworks.com Q: Can it be ball thrown v upwards, hitting ceiling(c) and bouncing back? 0 t Ball, released from rest at a certain height, hitting the floor and bouncing back
flipperworks.com(c) + v v2 g 0 t v2 -v1 v1 free fall ⇒ acceleration = gradient = -g
flipperworks.com Example 2: v/ s-1 m Velocity after 10 s = 0.80 x 10 = 8.0 m s-18.0 total distance travelled = area under graph = ½(20 +30) 8.0 t/ s = 200 m 0 10 30
flipperworks.com Example 3: The graph shows the variation with time of the velocity of a trolley, initially projected up an inclined runway.Velocity/m s-1 0.8 0.6 0.4 0.2 θ 0 Time/ s -0.2 -0.4 -0.6 -0.8
flipperworks.com (a) maximum distance = area under v-t graph between t = 0 & t = 2.5 s = ½ × 0.80 × 2.5 = 1.0 mVelocity/m s-1 0.8 0.6 0.4 Trolley reaches 0.2 max. distance 0 Time/ velocity = 0 when s -0.2 -0.4 -0.6 -0.8
flipperworks.com(b) a = gradient of v-t graph 0.00 - 0.80 = = - 0.32 m s-2 2.5 - 0.0 ∴ deceleration = 0.32 m s-2 Velocity/m s-1 0.8 0.6 0.4 0.2 0 Time/ s -0.2 -0.4 -0.6 -0.8
flipperworks.com (c) displacement = total area under graph = 1.0 + (-1.0) =0mVelocity/m s-1 0.8 0.6 0.4 0.2 +1.0 0 Time/ s -0.2 -0.4 -1.0 -0.6 -0.8
flipperworks.com(d) Trolley travels 1.0 m up the runway with uniform deceleration, stops momentarily at t = 2.5 s and then accelerates uniformly down the runway. Velocity/m s-1 0.8 0.6 0.4 0.2 0 Time/ s -0.2 -0.4 -0.6 -0.8
flipperworks.com Displacement Speed Velocity Acceleration Average speed Average velocityWORDS & TERMS KINEMATICS EQUATIONSGRAPHS
flipperworks.com Equations representing uniformly accelerated motion in a straight lineSuppose that a body is moving with constantacceleration a and that in a time interval t, itsvelocity increases from u to v and itsdisplacement increases from 0 to s . v u 0 t Since a = d v / d t ⇒ a = v - u t Hence v = u + at -------------- (1)
flipperworks.comSince velocity increases steadily, u+vaverage velocity, < v > = 2Recall: displacement, s = average velocity × timeThus, s = ½ (u + v) t ------------- (2)Substituting (1) into (2), s = ut + ½ at2 -------------- (3)
flipperworks.com v -uFrom (1), t = aSubstituting this into (2), u+v v -u s = × 2 aTherefore v 2 = u 2 + 2 a s -------- (4)
flipperworks.comKinematics equations for uniformlyaccelerated motion in a straight line :v = u +ats = ½ (u + v) t recall & derives = u t + ½ a t2v2 = u2 + 2 a sSince u, v , a and s are vector quantities,their directions must be taken intoaccount when solving problems.
flipperworks.com Sign Conventions Eg. A ball is released from a certain height. Starting+ position s is -ve a is -ve v is -ve