02 Kinematics Lecture Notes Soln for Worked Examples 2023

02 Kinematics Lecture Notes (Solutions for Worked Examples)

Example 1 A car travels 3.0 km due east from a point O to a point A for 2.0 minutes where it stops at a traffic light at A for 4.5 minutes. It then continues 4.0 km due north of A to point B for another 3.5 minutes. Find the Note that as velocity is a vector quantity, it is necessary to find the direction for average velocity a) average speed b) average velocity of the car b) average velocity of the car

Example 2 Displacement Velocity Acceleration A (speeding up) B (slowing down) C (speeding up) D (slowing down) E (slowing down) + + + + + - + - - - - + - + -

Example 3 Displacement = A – B + C Distance = A + B + C

Example 4 (a) Since s, v and a are vectors, we need to choose a positive sign convention before drawing the graphs. In this example, we will take upwards as positive. All axes of the graph must be labelled as far as possible. To determine maximum height (maximum displacement): A student throws a ball vertically upwards at a speed of 20 m s-1 from level A. Neglect air resistance. Note: In the video, I solved this problem without the use of kinematics equations (instead using a more graphical approach). Both methods will give the same answer – as expected!

Example 4 A student throws a ball vertically upwards at a speed of 20 m s-1 from level A. Neglect air resistance. 20.3 m 20 m s-1 - 20 m s-1 - 9.81 m s-2

Example 4 (b) When the ball returns to the hand, its displacement is zero. Taking upwards as positive:

Example 5 A motorist traveling at 13 m s-1 approaches traffic lights, which turn red when he is 25 m away from the stop line. His reaction time is 0.70 s. If he brakes fully such that the car slows down at a rate of 4.5 m s-2, On which side of the stop line will he stop, and how far from the stop line will he stop? Distance travelled before reacting = 13 x 0.70 = 9.1 m Distance travelled during deceleration: Taking the direction of initial velocity to be positive ( ) 22 2 2 2 2 0 13 18.8 m2 2 4.5 v u as vus a =+ −−= = = − Total distance travelled = 9.1 + 18.8 = 27.9 m Thus, he stops (27.9 – 25) = 2.9 m beyond the stop line.

Example 6 Considering the forces on the box, a) Net force: the component of weight of box parallel to the slope b) The acceleration is not g. It is not in free fall. c) Since normal contact force and weight are constant, the resultant force acting on the box is constant. Hence, the acceleration is constant. Therefore, the equations of motion can be applied in case 1.

Example 6 Considering the force on the feather, a) Net force: Weight of feather, W mg ma a g= → = b) The acceleration is g. It is in free fall. c) The acceleration is constant. Hence, the equations of motion can be applied in case 2.