Download presentation

Presentation is loading. Please wait.

Published byDavid Thomson Modified over 3 years ago

2
1 Using Kinematic Equations 1. Write down the symbols, values and units (in SI) of given quantities 2. Write down the symbol of the quantities required 3. Select the equation that contains all of the symbols in 1. and 2. above e.g. A stone is released from a height of 20.0 m above the ground. Neglecting air resistance and using the acceleration due to gravity as 9.81 ms -2, find the velocity with which the stone will hit the ground. v f 2 = v i 2 + 2ad v f 2 = 0 2 + 2 x 9.81 x 20 v f = 392 = 19.8 m s -1 vivi = 0 from rest d= 20.0 m a= 9.81 ms -2 vfvf = ? v f 2 = 392

3
© John Parkinson 2

4
3 Distance travelled - d Time taken - t Velocity - v v= d t d v | t Velocity = Speed in a Specified Direction Constant Velocity

5
4 N 100 m in 4 seconds Distance travelled = ?100 m Displacement = ? 100 m to the East Speed = ?Speed = 100/4 = 25 m s -1 Velocity = ? Velocity = 25 m s -1 to the East

6
5 Displacement / time graphs Constant velocity Displacement d Time t What will the graph look like? Gradient = ? Δt Δd Velocity

7
6 Displacement - d Time - t What about this graph? A body at rest Displacement d Time t And this graph? The gradient is …….? increasing Δd Δt The body must be ……..? accelerating

8
7 1 3 2 AVelocity/time graphs Velocity - v Time - t Velocity - v Time - t This body has a constant or uniform ………? acceleration Δv Δt The gradient = ? the acceleration 1 = …… ? Uniform acceleration 2 = …… ? Constant velocity 3 = …… ? Uniform retardation [deceleration] Area under the graph = A = …….. ? Distance travelled

9
8 Velocity v (ms -1 ) Time t(s) 30 2050 80 Question The graph represents the motion of a tube train between two stations Find 1.The acceleration 2.The maximum velocity 3.The retardation 4.The distance travelled 1. The acceleration = the initial gradient = 30÷20 = 1.5 m s -2 2. The maximum velocity is read from the graph = 30 m s -1 3. The retardation = the final gradient = -30 ÷ [80-50] = m s -2 4.The distance travelled = the area under the graph =½ x 20 x 30 + x + ½ x x = 1650 m

10
9 Velocity Time What might this graph represent? Can you draw an acceleration time graph for this motion? Terminal Velocity Acceleration Time 9.81 m s -1

11
Find the acceleration at each part Time (s) 0 1 2 3 4 36 91215 5 -2 -3 -4 5 18 17 Velocity (ms -1 ) 0-3 s: a =1 ms -2, accelerating 3-5 s: a = 0 ms -2, constant velocity 6-9 s: a = 0 ms -2, v=0, rest 5-6 s: a = -3 ms -2 decelerating 9-12 s: a = - 1.3 ms -2 accelerating to –ve direction 12-15 s: a = 0 ms -2, constant velocity, -ve direction 15-17 s: a = 2 ms -2, decelerating

12
Find out the displacement 0-6 s: d 1 =(6+2)x3/2=12m, to +ve direction Time (s) 0 1 2 3 4 36 91215 5 -2 -3 -4 5 18 17 + – Velocity (ms -1 ) 6-9 s: rest, d 2 = 0 9-17 s: d 3 =- (8+3)x4/2=-22m, to -ve direction Total displacement d = d 1 + d 2 + d 3 = -10m, to -ve direction Total distance = 34m Average speed = 34/17 = 2ms -1 Average velocity = -10/17 = -0.59 ms -1

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google