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 Velocity. (m/s) 10 12 14 -6 Time, t(s) a) Determine the acceleration of the object at 1.0s. b) Determine the average velocity of the object between 4.0 and 8.0 s. c) Over what time interval(s), 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: Ar-600km at 25.0 N of W, followed by Ars = 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. marks)

Answer 1

Using formula for acceleration from vt graph (slope of the graph)

Displacement from vt graph (area under graph)

Average velocity can be calculated easily

is (as from obsequation of giren vat graph slope of vot graph goaph is constaint for t= 0 to 4 = 2 sec that means slepe (= acceleration} is cons for this duration slops = 8-5 V₂ - VI titi 11 بازم 2-0 Slope = a = 1,5 1.5 m/s² at ta los To find (b) Again from above vet graph average velocity. we need to fit calculate graph from t = 4sto += 85 it can be divided in two parts from t=4sto t=65 and from t=6s to t=8s area under tre a = X(6-4) (8-0) Area į (8-4) 6-3-6) 2x 8 2X-3 2 =8m = -3 Total displacement 8+(-3) = 8-3 = 5m

average velocity Vo = total diophoement total time 5=1.25 ml 5 4 = 1.25 m) { } (6 Yes, in a vet graph where ever in negative slope that means body 3 Las -ve acceleration deaccelerating (slowing down? a ve slo on observation we can see in graph object has slope from/t = 6s to t= 105 that means for this duration body is deacceleration (d) yes. in ou vot graph where ever graph touch zero that shows body has zero welocity at that point in this graph alt= 66 object has zoro velocity = atresti t 125
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