2.2 Equations of uniformly accelerated motion Skid marks Uniformly accelerated motion Derivation of equations of motion Check-point 5 1 2 Book 2 Section 2.2 Equations of uniformly accelerated motion

Book 2 Section 2.2 Equations of uniformly accelerated motion Skid marks In an accident, drivers brake their cars hard, skid marks are left on the road. The police will measure the length of the skid marks. Why? To estimate the speed of car before it brakes. Book 2 Section 2.2 Equations of uniformly accelerated motion

1 Uniformly accelerated motion Consider an object accelerating uniformly from initial velocity u to final velocity v over time t. Book 2 Section 2.2 Equations of uniformly accelerated motion

1 Uniformly accelerated motion Average velocity (v ) = u + v 2 Displacement = area under v-t graph (i.e. a trapezium) = (u + v)t 2 Book 2 Section 2.2 Equations of uniformly accelerated motion

2 Derivation of equation of motion Consider a general case: uniform acceleration = a initial velocity = u final velocity = v time = t Four important equations for uniformly accelerated motion can be derived. Book 2 Section 2.2 Equations of uniformly accelerated motion

2 Derivation of equation of motion By the definition of acceleration, a = v – u t ……(i) Re-arrange (i), v = u + at ……(1) Area under the graph equals displacement s, 1 2 s =  (u + v)  t ……(2) Book 2 Section 2.2 Equations of uniformly accelerated motion

2 Derivation of equation of motion Sub (1) into (2), s = ut + 1 2 at 2 …(3) v – u Sub t = into (2), a 1 2 v – u a s =  (u + v)  2as = (v + u)  (v – u) v 2 = u 2 + 2as …(4) Book 2 Section 2.2 Equations of uniformly accelerated motion

2 Derivation of equation of motion The four equations obtained are: v = u + at 1 2 s = (u + v)  t ut + at 2 v 2 = u 2 + 2as They are called the equations of motion. Book 2 Section 2.2 Equations of uniformly accelerated motion

2 Derivation of equation of motion When applying the four equations, 1 acceleration must be constant; 2 the signs of s, u, v and a should be consistent with the defined +ve direction. Example 9 Velocity of a powerboat Book 2 Section 2.2 Equations of uniformly accelerated motion

2 Derivation of equation of motion Example 10 Road test on brake Book 2 Section 2.2 Equations of uniformly accelerated motion

2 Derivation of equation of motion Stopping distance, thinking distance and braking distance Example 11 Book 2 Section 2.2 Equations of uniformly accelerated motion

Book 2 Section 2.2 Equations of uniformly accelerated motion Check-point 5 – Q1 True or false: A car accelerates from t1 to t2 non-uniformly from v1 to v2 over distance d. The average velocity of the car is v1 + v2 2 . (T / F) Book 2 Section 2.2 Equations of uniformly accelerated motion

Book 2 Section 2.2 Equations of uniformly accelerated motion Check-point 5 – Q2 Maria (1.5 m tall) is standing under an apple tree. An apple 2.5 m right above her falls down with an acceleration of 10 m s–2. Find the time the apple takes to hit her. Book 2 Section 2.2 Equations of uniformly accelerated motion

Book 2 Section 2.2 Equations of uniformly accelerated motion Check-point 5 – Q2 ? u = ________ t = ________ 10 m s–2 v = ________ ? a = ________ 2.5 – 1.5 = 1 m s = ______________ v = u + at 1 2 s = (u + v)  t ut + at 2 v 2 = u 2 + 2as  1 = 0 + 1 2 (10)t 2 t = t = 0.447 s Book 2 Section 2.2 Equations of uniformly accelerated motion

Book 2 Section 2.2 Equations of uniformly accelerated motion Check-point 5 – Q3 (a) Compared with stopping on a dry road, a car stopping on a wet road will have a longer ___________ distance and ___________ distance. braking stopping (b) If the deceleration of the car is halved, the stopping distance (will/will not) be doubled. Book 2 Section 2.2 Equations of uniformly accelerated motion

Book 2 Section 2.2 Equations of uniformly accelerated motion Check-point 5 – Q4 Blackbird (Lockheed SR-71) was the fastest jet in the world. Max. speed = 894 m s–1 Max. acceleration = 3.75 m s–2 Estimate the shortest distance travelled when it accelerates from rest to its top speed. Book 2 Section 2.2 Equations of uniformly accelerated motion

Book 2 Section 2.2 Equations of uniformly accelerated motion Check-point 5 – Q4 ? u = ________ t = ________ v = ________ 894 m s–1 a = ________ 3.75 m s–2 ? s = ________ v = u + at 1 2 s = (u + v)  t ut + at 2 v 2 = u 2 + 2as  8942 = 0 + 2(3.75)s s = 107 000 m Book 2 Section 2.2 Equations of uniformly accelerated motion

Book 2 Section 2.2 Equations of uniformly accelerated motion The End Book 2 Section 2.2 Equations of uniformly accelerated motion