1.3 velocity

1.3 velocity

• 1.
  LECTURE 4 Acceleration IB PhysicsPower Points Topic 2 Kinematics www.pedagogics.ca
• 2.
  Defining Acceleration Acceleration isused commonly to mean “speeding up” Acceleration in physics has a very specific meaning. Acceleration is a measure of how quickly velocity changes. To change velocity an object must undergo an acceleration. Recall – velocity is a vector. A change in velocity can be a change in speed OR direction OR both speed and direction. Example 1: a car travels around a corner bend in the road at a constant speed of 45 km/h. Is the car’s speed changing? No (read the question). Is the car’s velocity changing? Yes, because the direction of travel is changing. Therefore this car is also accelerating.
• 3.
  Defining Acceleration Example 2:The driver of a car travelling along a straight road at 45 km/hr sees the light ahead turn red. The driver slows and stops at the intersection. Is the car accelerating? Yes. The car’s velocity is changing. In this case, the car is slowing down. This would be a negative acceleration. Mathematically acceleration can be calculated by: 2 1v vv a t t      In words, acceleration is the change in velocity over a time interval DIVIDED by the time interval. Unit for acceleration is ms-2 or m/s2 or m/s/s
• 4.
  Quick problem #1:A skater changes her velocity from 3.5 ms-1 [N] to 2.5 ms-1 [S] in 3.4 s. What is her acceleration? Quick problem#2: A car travelling at 24 km/h speeds up at a rate of 0.52 ms-2 for 3.7 s. What is the final speed? 2 1 2.5 3.5 3.4 v v a t       -2 -2 1.8 ms [ ] = 1.8 ms [ ] N S   2 1v v a t   2 6.67 (0.52)(3.7)v   -1 2 8.6 ms or 31 km/hrv 
• 5.
  Velocity time graphstell us how velocity changes. The slope of a velocity time graph indicates acceleration. t v [up] This graph shows how the velocity of a falling rock changes. What would the slope be?
• 6.
  Distance Travelled byObjects Moving at Constant Velocity We can calculate the change in position of an object travelling at constant speed very easily. s v t    Distance = velocity x time DO NOT USE THIS EQUATION FOR ACCELERATING OBJECTS!! v t t The graph shows a situation where the above equation applies. How would you determine Δs from just the graph?
• 7.
  Distance Travelled byObjects Moving at Constant Velocity From this observation we can make a very important general conclusion: THE AREA UNDER A VELOCITY-TIME GRAPH REPRESENTS DISPLACEMENT (distance). s v t    v tt Δs = v x Δt would be the same calculation as determining the area under the graph for the given time interval. area v t  
• 8.
  Distance Travelled byAccelerating Objects v2 Δt For a uniformly accelerating object the same principle can be applied. 1 2 displacement area area b h    2 2 1 1 Recall 2 So s t v v v a t       Initial velocity is zero 2 1 1 When 2 Substitution gives s a t v zero   
• 9.
  Distance Travelled byAccelerating Objects When there is an initial velocity the graph looks like this… The total area under the curve gives us the displacement. This is the combined areas of the triangle and the rectangle. 2 1 1 2 Mathematically s v t a t     v2 t v1 21 2 s a t   1s v t    Remember this formula!
• 10.
  Distance Travelled byAccelerating Objects Summary so far ……. 1. The slope of a velocity vs time graph indicates acceleration. 2. The area under a velocity vs time graph indicates displacement. 3. To find displacement of an accelerating object we can use the formula:      2 1 1 2 s v t a t
• 11.
  Distance Travelled byAccelerating Objects We can also use this graph to derive a different formula. This time use a slightly different calculation for the triangle:                   1 2 1 2 1 1 1 2 1 ( -v ) 2 v 2 2 2 Mathematically s v t t v v t t s v t v v s t v2 t v1    2 1 1 ( ) 2 s t v v 1s v t   
• 12.
  Distance Travelled byAccelerating Objects The critical equations for uniform acceleration are: 1. 2. 3. We can use substitution to produce a final useful equation. 4.         2 1 or v vv a v a t t t      2 1 1 2 s v t a t     1 2 2 v v s t       2 2 2 1 2v v a s
• 13.
  In your databooklet
• 14.
  Examples A car travellingat 13.0 ms-1 coasts for 17 s decreasing its velocity to 2.4 ms-1. How far did the car coast?             1 2 2 13 2.4 17 130 m 2 v v s t s Identify variables. 1v2v t Identify unknown  displacement. Choose equation, substitute and solve:
• 15.
  Examples What was theinitial velocity of a rocket that accelerates upwards at 0.175 ms-2 to reach a velocity of 36.7 ms-1 over an altitude change of 1250 m?                2 2 2 1 2 1 2 2 1 -1 1 2 2 36.7 2(0.175)(1250) 30.2 ms v v a s v v a s v v Hint: the equation used is very useful when you are not given time.
• 16.
  Freefall An object issaid to be in freefall when the only force acting on it is gravity. As we ignore air resistance in most calculations, any falling object is in freefall. The acceleration of a freefalling object depends on the force of gravity. On Earth … -2 9.81 ms [down]gravitya  It is conventional in physics to use the letter “g” to represent this special acceleration. It is also conventional to use [up] as the positive direction in the reference frame so in most cases we substitute a = g = -9.81 ms-2.
• 17.
  Examples A ball istossed upwards with an initial velocity of 15.6 ms-1 from the roof of the school. The ball reaches maximum height and falls back down, missing the edge of the roof and hitting the ground below. The velocity at impact on the ground was 26.5 ms-1 [down]. How long was the ball in the air?       2 1 2 1 26.5 15.6 4.29 s 9.81 v v a t v v t g           