HCI Assignment 03 Answer Kinematics

Kinematics 1 Name: Index No: Date: Class: Assignment 03: Kinematics (1) Four different mice (labelled A, B, C, and D) ran the equilateral triangular maze shown in the diagram below. They started in the lower left corner and followed the paths of the arrows. The times they took are also shown in the diagram. State which mouse/mice fits/fit the following description. (a) This mouse had the greatest average speed. [1] B. Total distance travelled per unit time is the greatest. A: s/2 B: 2s / 2 = s C: 2s / 4 = s/2 D: 3s / 4 (b) These two mice had the same greatest change of displacement. [1] A and C. Both have the same change of displacement (i.e. their final positions are at lower right corner). B and D have zero change of displacement. (c) These two mice had an average velocity that points to the right. [1] A and C. Both have the same change of displacement which points to the right. [For C, use addition of vectors to get change of displacement.] (d) This mouse had the greatest average velocity. [1] A. For the same change of displacement for A and C, A takes a shorter time. HWA CHONG INSTITUTION Sec 3 Physics

Kinematics 2 (2) Randall was taking Lucky, his dog, for a walk early in the morning towards the rising Sun, when he decided to take note of their displacement with respect to time. He plotted the following graph as shown below. (a) Calculate the velocity of Randall and Lucky during the first 2.0 minutes, in m s -1. [2] velocity = (30 – 0) / [(2.0 – 0) x 60] [1] = 0.25 m s-1 [1] [only need magnitude] (b) Calculate the total distance covered by Randall in 7.0 minutes. [1] total distance covered = 90 + 60 = 150 m [1] (c) What is the change of displacement of Randall and his dog at the end of the 7.0 min walk? [1] change of displacement = 30 m, towards the rising Sun or due East [state direction] [1] (3) The diagram shows a graph of the variation of the velocity v with time t for an object moving in a straight line. (a) Describe the acceleration of the object from t = 4 s to t = 16 s. [1] Acceleration is decreasing and positive. [DO NOT write “accelerate at a decreasing rate” because acceleration is already a rate. Instead, decreasing acceleration has the same meaning as velocity increasing at a decreasing rate.] displacement / m 0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 time / min v / m s-1 t / s

Kinematics 3 (b) State the velocity of the object when its acceleration is (i) at its maximum value, velocity = _0 m s-1 ____ (ii) zero. velocity = _20 m s-1___ [2] (c) Calculate the rate of change of velocity of the object during the last part of its motion. [1] a = (0 - 20) / (24 - 20) = - 5 m s-2 Rate of change of velocity is - 5 m s-2. (4) The diagram shows how the velocity of a body moving in a straight line varies over a period of time. Take positive velocity to be the rightward direction. (a) Identify and state the region(s) for each motion of the body below. [6] Region(s) (i) speed increases at a constant rate AB (const +ve acceleration in +ve velocity direction) EF (const -ve acceleration in -ve velocity direction) (ii) non-zero constant velocity BC (zero acceleration in +ve velocity direction) FG (zero acceleration in -ve velocity direction) (iii) speed decreases at a constant rate CD (const -ve acceleration in +ve velocity direction) GH (const +ve acceleration in -ve velocity direction) (iv) zero velocity DE (stationary) (v) constant positive acceleration AB (v  as it moves in +ve velocity direction) GH (v  as it moves in -ve velocity direction) (vi) constant negative acceleration CD (v  as it moves in +ve velocity direction) EF (v  as it moves in -ve velocity direction) (b) Describe the acceleration of the body over the region marked EF on the graph. [1] Constant negative acceleration OR Constant acceleration (of 2 m s-2) to the left (in the leftward direction) [acceleration = (-10 - 0) / (40 - 35) = - 2 m s-2] [Wrong: Constant deceleration. It is speeding up in the opposite direction, not slowing down.] A G H velocity / m s-1 time / s

Kinematics 4 (c) Explain how the distance travelled by the body and the change of displacement of the body is determined from the graph over a duration of 50 s. [2] Distance travelled is determined by finding the sum of area ABCD and area EFGH. [1] Change of displacement is determined by finding the difference of area ABCD and area EFGH. [1] Since the graph shows the body changing its direction of motion (from E to H, object is moving to the left), this means the distance travelled by the body is greater than its change of displacement. (d) Determine the average velocity over the region AF. [2] average velocity over AF = change of displacement / total time = [½ (20)(10 + 25) – ½ (5)(10)]/40 [1] = [350 – 25] / 40 = 8.1 m/s (to 2 s.f.), to the right [Give magnitude and direction.] [1] (5) A stone is projected almost vertically upwards at 20 m s −1 from the edge of a cliff. It finally lands on the ground at the base of the cliff. The sequence diagram below shows the position of the stone at one - second intervals. Position 0 is just after projection and position 5 is just before landing. Gravitation al acceleration is taken as 10 m s−2 and air resistance is ignored. (a) Draw in a vector next to positions 1, 3, 4 and 5 to represent the instantaneous velocity at that stage of the motion. The vector at position 0 has been drawn for you. Pay attention to the direction and relative lengths of the vectors, and label them with their magnitudes, in m s-1. [2] Any 2 correct vectors (direction and relative vector length): [1] All 4 correct vectors (direction and relative vector length): [2]

Kinematics 5 (b) On the diagram below, draw a velocity –time graph to represent the motion of the stone. Label the stages representing the upwards motion and the downwards motion and also label the highest point of the motion. Take positive velocity to be upward direction. [2] Correct graph: [1] Correct labels: [1] (c) What does the gradient of the graph represent? [1] Stone’s acceleration / Acceleration due to gravity / Acceleration of free fall [Apply to context instead of just writing acceleration.] (d) Determine the height of the cliff, in m. [2] Height of cliff = change of displacement from t = 4 s to t = 5 s = area of trapezium = ½ (5 - 4)(20 + 30) [1] = 25 m [1] highest point

Kinematics 6 (6) A straight road links point A to point B. Car P moves from point A to point B while car Q moves from point B to point A as shown in the diagram below. The diagram also shows the speed -time graph of the two cars. (a) Describe the acceleration of car P from time = 0 s to time = 50 s. [2] From time = 0 s to time = 10 s, car P undergoes constant positive acceleration (or constant acceleration of 2.5 m s-2). [1] From time = 10 s to time = 50 s, its acceleration is zero. [1] [X: 0 acceleration. Write out the term.] (b) The two cars met at time = 25 s. (i) What is the numerical difference in the speed between car P and car Q at the time when the cars met? [1] 25 – 15 = 10 m s-1 [Necessary