HCI 02 Kinematics Solutions (Discussion Questions)

1 Tutorial 2A: Kinematics Discussion Questions (Suggested Solution) D1 (a) Yes. An object in uniform circular motion moves at constant speed but its direction of motion is changing all the time (its velocity is changing) and hence it has acceleration. Circular Motion will be covered in greater detail in chapter 6. (b) No. The ma gnitude of the velocity is the speed of an object. If the velocity is constant, both magnitude and direction of velocity have to be constant. Hence, speed cannot change. (c) Yes. Projectile motion of a projectile under free -fall, with no air resistance . The direction of motion (velocity) changes with time. The acceleration, g, is constant. (d) Yes. The object can be instantaneously at rest and the next moment its velocity increases or decrease. Eg an object thrown vertically upwards and at its highest p oint, it is instantaneously at rest but it is still accelerating downwards with g. (e) Yes. That happens when the object is slowing down. (f) No. Consider an object initially moving with some vel ocity then resting for some time then continuing to move with some velocity. The average velocity is not zero as net displacement is not zero but during this interval it was at rest at some point. D2 At max height, velocity is zero and gradient at that point is equal to acceleration of free fall. Also, point A is the point it first hits the ground. NOTE: area of triangle ABC should be equal size to area of triangle CD since the ball rises and falls through the same distance after the first bounce. Ans: C D3 The area under velocity-time graph gives the displacement. Both objects have the similar acceleration rates throughout. You may sketch the displacement versus time graph for both objects. The displacement for P is obviously much greater. In fact, the area under velocity-time for Q suggests that the displacement of Q is zero at the end of time t2. To compare the distance travelled by P and Q, we could make use of the area under velocity-time graphs Object P: 2 2 vt

2 Object Q: ( )211211 222 2 2 2 2 2 4 ttt vtvv  −  +  =   . The distance travelled is also different. Ans: A D4 The area under velocity-time graph gives the displacement of the cars. At t = 0, both are at the same position where Car X takes over. At t = T, both cars must also be at the same position so that Car Y could take over. The displacement of Car X at t = T is P + Q + R and that of car Y is S + Q + R. The displacements must be the same. Hence, P is equal to S. Ans: A D5 Area under the graph represents change in velocity. At point C, the area (from start to point C) is greatest. Beyond point C there is a negative change in velocity, which means the velocity is decreasing. Ans: C D6 Method I (equations of motion) Let t = 0 be when the ball passes by light gate 1. At light gate 2, 2 1 1 2s ut at=+ (1) At light gate 3, 2 12 1(2 ) (2 )2s s u t a