# SPH4U – Grade 12 Physics Unit 3

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SPH4U – Grade 12 Physics Unit 3
Kinematics Review SPH4U – Grade 12 Physics Unit 3

Review - Motion The study of motion is called kinematics.
In physics we study scalar and vector quantities. Scalar quantities do not have a direction. Ex. Speed = 2 km/hr Vector quantities have both a magnitude and a direction. Ex. Velocity = 3 m/s [South]

Review - Motion Position is the distance with a direction from some reference point. Displacement is the change is position. Velocity is the rate of change of displacement. Instantaneous velocity is the velocity at a particular instant in time. It is found by finding the slope of the tangent at that point to a displacement-time graph.

**Most motion is non-uniform. **
Uniform Motion Recall: Uniform motion is movement in a straight line at a constant speed. **Most motion is non-uniform. ** The position-time graph for uniform motion will be a straight line with a positive slope.

Review – Motion Graphs On a displacement time graph (p-t) or (d-t), the slope represents velocity.

Uniform Motion Non-uniform motion, by default, is accelerated motion. Accelerated motion occurs when we have a change in speed or direction or both. Uniformly accelerated motion is when an object is travelling in a straight line and changes its speed uniformly with time.

Uniform Motion Uniformly accelerated motion (also known as uniform acceleration) will have a straight line on a v-t graph and a flat straight line on an a-t graph. ** In this course, we only examine things that have uniform acceleration.

Uniform Motion Uniformly accelerated motion

Review – Motion Graphs On a v-t graph, the slope represents acceleration because acceleration is the rate of change of velocity.

Practice: 1) In the graph below, explain what is happening to the objects motion for each time interval.

Practice: 1) In the graph below, explain what is happening to the objects motion for each time interval. Not moving Constant velocity Away from origin Constant velocity toward origin Accelerating away from origin

Practice: 2) Using the p-t graph given, construct the v-t graph. Determine the average velocity between 0 and 6 seconds. Determine the instantaneous velocity at 3 seconds.

2) Using the p-t graph given, construct the v-t graph
2) Using the p-t graph given, construct the v-t graph. Determine the average velocity between 0 and 6 seconds. Determine the instantaneous velocity at 3 seconds. Practice: Average velocity = 5/6 m/s = 0.83 m/s Inst. Velocity = slope of straight line = 1 m/s v-t graph 1 velocity (m/s) -2 time (s)

Practice: 3) Using the v-t graph given, sketch the a-t graph.

Practice: 3) Using the v-t graph given, sketch the a-t graph.
6 Acceleration (m/s2) 2.6 -4 time (s)

Review – Equations of Motion
Recall the following important equations of motion (you can find these in Table 1 on pg. 18)

Practice: 4) A man starts from rest and then runs north with a constant acceleration. He travels 120m in 15 seconds. Calculate his acceleration.

Practice: 4) A man starts from rest and then runs north with a constant acceleration. He travels 120m in 15 seconds. Calculate his acceleration.

Review – Free Fall Free fall is the motion of an object falling to the earth’s surface with no other force acting on it other than gravity. The average acceleration on the Earth due to the Earth’s force of gravity is This is also called the gravitational field intensity.

Review – Projectiles Projectile motion is motion with a constant horizontal velocity combined with a constant vertical acceleration caused by gravity. A projectile is an object that moves through the air, along a trajectory without a propulsion system.

Review – Projectiles Recall the following equations that are useful for projectile problems:

Practice: 5) A projectile is launched with an initial speed of 14.5 m/s at an angle of 35.0º above the horizontal. The object lands at the same height from which it was launched. What is the projectile’s maximum height?

Practice: 5) A projectile is launched with an initial speed of 14.5 m/s at an angle of 35.0º above the horizontal. The object lands at the same height from which it was launched. What is the projectile’s maximum height?

Relative Motion When we observe motion, we are always measuring it with reference to some particular frame. Relative motion is motion observed from a specific perspective or frame of reference. The velocity of an object relative to a specific frame of reference is called relative velocity.

Relative Motion We can solve relative velocity questions using the following equation: = velocity of A moving relative to C = velocity of A moving relative to B = velocity of B moving relative to C

Example Suppose the school bus moves at a velocity of 40 km/h [E], relative to the ground. A student rolls an apple down the middle of the bus at a velocity of 2 km/h [E] relative to the student. What is the velocity of the ball relative to the ground?

Example Suppose the school bus moves at a velocity of 40 km/h [E], relative to the ground. A student rolls an apple down the middle of the bus at a velocity of 2 km/h [E] relative to the student. What is the velocity of the ball relative to the ground? = velocity of the bus moving relative to the ground = velocity of the apple moving relative to the bus = velocity of the apple moving relative to the ground

Example Suppose the school bus moves at a velocity of 40 km/h [E], relative to the ground. A student rolls an apple down the middle of the bus at a velocity of 2 km/h [E] relative to the student. What is the velocity of the ball relative to the ground? = 40 km/h [E] = 2 km/h [E] = ?

Practice: 6) A child rolls a ball across a boat at a velocity of 4 m/s [N] (relative to the boat). If the boat is travelling at 6m/s [W] relative to the shore, what is the velocity of the ball relative to the shore?

Practice: 6) A child rolls a ball across a boat at a velocity of 4 m/s [N] (relative to the boat). If the boat is travelling at 6m/s [W] relative to the shore, what is the velocity of the ball relative to the shore? vbB = 4m/s [N] vbs = vbB + vBs vBs = 6m/s [W] vbs = 4 [N] + 6 [W] (use pathagorean theorem) vbs = 7.2 [N 56º W] vbs = ?

Homework Read sections Supplement your notes with these readings. Complete the Self-Quiz on page. 53 On pg. 55 Answer Questions #36, 38, 48, 54, 55, 59, 67, 75, 77