HCI 02 Kinematics Solutions (Self-Review Questions)

1 Tutorial 2A: Kinematics Self-Review Questions (Suggested Solution) S1 (a) Change in speed = final speed – initial speed = (30 – 10) m s-1 = 20 m s-1 (b) Change in velocity Change in velocity = vfinal – vinitial = vfinal + (– vinitial) Magnitude of the change in velocity Using cosine rule, v = √(102 + 302 – 2(10)(30) cos 60o) = 26.5 m s-1 Direction of the change in velocity Using sine rule, sin α sin 60o 10 = 26.5 α = 19.1o The change in the velocity is 26.5 m s-1 at an angle of 79.1o anticlockwise from the horizontal (c) Average speed = total distance travelled total time taken = (10 x 5.0) + (30 x 2.0) (5.0 + 2.0) = 15.7 m s-1 (d) Average velocity Using cosine rule, ∆𝑠⃗ = √(602 + 502 – 2(60)(50) cos 120o) = 95.4 m Magnitude of the average velocity, |v|= 95.4 / 7.0 = 13.6 m s-1 Using sine rule, β = 33.0o The average velocity is 13.6 m s-1 at 33.0o to the initial velocity. S2 (a)To find total displacement AC, =ABs 6.4960 3585 = km Using cosine rule, )4590cos(2222 oo BCACBCABAC sssss +−+= )135cos(1306.4921306.49 222 o ACs −+= 169=ACs km Using sine rule, N A C 35 min, 85 km h-1 2 h, 130 km Stop for 15min B 

2 o ACBC ss 135sinsin = o0.33= (b)Average velocity = 6.59 260 15 60 35 169 = ++ km h-1, o033. east of north Points to note: Students are expected to be able to recall the cosine and sine rule as these formulas are not given in any Physics test/exam. Answers should only be written in terms of North, South, East, West if such terminology were used in the question in the first place. S3 (a)Time taken = Distance travelled / speed = 14304 04 .. ).( == v R s (b) average velocity total change in displacement total time= = 552143 0404 .. .. =+ m s-1 A to B (c) change in velocity, 0.8=−= if vvv  m s-1, upwards At A, iv At B, fv For v S4 (a)average velocity between t = 0.0s and 3.0s, total change in displacement total time= = 83.00.00.3 0.45.1 −=− − m s-1 (b) At t = 3.0s, instantaneous velocity = 𝑑𝑥 𝑑𝑡 = gradient of tangent line at t = 3.0s = 67.02.50.0 0.05.3 −=− − m s-1 Exam skills: The 2 points chosen as coordinates for the gradient of the tangent line must be at least ½ the span of the graph. Points to note: A common misconception is students has the wrong concept of velocity = 𝑥 𝑡 and think that instantaneous velocity = 𝑥 𝑡 = 1.5 3 = 0.5 m s-1 . S5 Gradient of d-t graph is v-t graph. Note that at t = 0, the v is zero and hence the gra