Question

5. The following graph represents the velocity of an object, moving along a horizontal line as a function of time. Express all answers to two significant figures. East is the positive direction and West is negative. (6 marks) Velocity Vs. Time 10 8 6 Velocity. (m/s) 10 12 14 -2 -6 .8 Time, t, (s) a) Determine the acceleration of the object at 1.0. b) Determine the average velocity of the object between 4.0 and 8.0 s. c) Over what time intervalls), if any, is the object decelerating (slowing down)? d) At what time(s), if any, does the object remain at rest? 6.A bus undergoes two successive displacements; 17; - 60.0 km at 25.00 N of W, followed by Ar = 50.0 km W. Use a scaled diagram to determine the net displacement of the bus. Clearly state the scale and the lengths of the vectors. (5 marks)

Answer 1

a. Acceleration of the object at t = 1 s is equal to the slope of the graph at t = 15 8-5 2-0 = 1.5 x m/s2 b. Average velocity of the object between 4 to 8s is equal to the area of the graph between 4 to 8 s divided by the time interval 4 to 8 S. Area above time axis where velocity is positive is considered as positive and area below time axis where velocity is negative is considered as negative Svdt (Area) 4 αν 8-4 dt 4 {(t)-( *(8) (6-4). * (3) (8-6) 8-6)} Yav 4 Vav = 1.25 x m/s C. Acceleration is represented by the velocity time graph. Object is slowing down when velocity and acceleration are in opposite directions i.e. the sign of velocity and acceleration is opposite. According to this concept object is slowing down in interval 4 to 6 and 10 to 12 seconds d. Object momentarily comes at rest at the time when it's instantaneous velocity becomes zero i.e. at t = 6 second 6.

Tworth 1 cm -10km 40 50 km 472 or 60km 20 AY 14° West 25. -20 -100 -80 -60 -40 -20 -40 Ar 6km Ar2 = 5km Net displacement = 10.7cm = 107km direction 14° N of W