--- title: "What is acceleration, exactly?" subtitle: "Year 10 Physics · Newton's laws · Lesson 3" --- # What is acceleration, exactly? **Name:** {{field-24}} **Date:** {{field-15}} **Class:** {{field-15}} > **Key question:** A Formula 1 car reaches 100 km/h in under 3 seconds. A train takes about 40. Both end up fast; one accelerates far more. Fast and accelerating are not the same thing. :::teacher-note **In one sentence:** Students define acceleration as the rate of change of velocity, calculate it from data and graphs in m/s², and treat it as a vector in the direction of the net force (AC9S10U05, node 3, a hinge node). **Students should be able to say:** I can define acceleration as the rate of change of velocity, in m/s². I can calculate acceleration from velocity and time data, and read it from a speed–time graph. I can explain that acceleration has a direction, the same as the net force; slowing down is acceleration too. **Width:** the unmarked parts are the floor every student does (Engage, Explore, Explain, Evaluate). Typical and Wider sections add contexts, not difficulty. Students work in the online tutorial alongside this sheet. ::: ## Engage :::keep-together **Which has the bigger acceleration: a car cruising at a steady 100 km/h, or a cyclist going from 0 to 20 km/h in 4 seconds?** ☐ Neither has any acceleration ☐ Both the same ☐ The cyclist, because their speed is changing ☐ The car, because it is much faster My prediction, and why: {{lines-3}} ::: :::teacher-note **What happens:** The cyclist. The car's speed isn't changing, so its acceleration is zero, however fast it goes. Acceleration is how quickly velocity changes. ::: ## Explore Try it on the board in the tutorial. Was your prediction right? What was different? {{lines-2}} ## Explain: a = Δv ÷ Δt :::keep-together ### From the data **The trolley went from 0 to 12 m/s in 4 s. What was its acceleration, in m/s²?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 3 a = change in velocity ÷ time = (12 − 0) ÷ 4 = 3 m/s². Its speed went up by 3 m/s every second. ::: :::keep-together ### On the graph **On the speed–time graph, where can you see the acceleration?** ☐ Where the line ends ☐ How high the line is ☐ The gradient (steepness) of the line ::: :::teacher-note **Answer:** The gradient (steepness) of the line A steeper line means the speed changes faster: bigger acceleration. A flat line means zero acceleration. ::: :::keep-together ### Units **What are the units of acceleration?** ☐ N ☐ m/s² (metres per second, per second) ☐ m/s ☐ km/h ::: :::teacher-note **Answer:** m/s² (metres per second, per second) ::: :::keep-together ### Another one **A cyclist goes from 2 m/s to 10 m/s in 4 s. Acceleration, in m/s²?** Answer: {{field-15}} ::: :::teacher-note **Answer:** 2 (10 − 2) ÷ 4 = 2 m/s². ::: :::keep-together ### Accelerating or not **Is it accelerating?** Groups: Accelerating, Not accelerating - A car cruising at a steady 100 km/h on a straight road: {{field-15}} - A bike slowing down for traffic lights: {{field-15}} - A runner rounding a bend at a steady speed: {{field-15}} - A trolley speeding up from rest: {{field-15}} - A parked car: {{field-15}} ::: :::teacher-note **Answer:** A car cruising at a steady 100 km/h on a straight road → Not accelerating; A bike slowing down for traffic lights → Accelerating; A runner rounding a bend at a steady speed → Accelerating; A trolley speeding up from rest → Accelerating; A parked car → Not accelerating ::: :::keep-together ### Which way **A trolley is moving to the right, and the net force on it points to the left. Which way does its acceleration point?** ☐ Nowhere: it has no acceleration while it still moves right ☐ Right, the way it is moving ☐ Left, the same way as the net force ::: :::teacher-note **Answer:** Left, the same way as the net force Acceleration always points the same way as the net force. Here that is backwards, so the trolley slows down. ::: :::typical ## Elaborate: slowing down :::keep-together ### Negative acceleration **It went from 8 m/s to 0 in 4 s. What was its acceleration, in m/s²? (Use a minus sign for slowing down.)** Answer: {{field-15}} ::: :::teacher-note **Answer:** -2 (0 − 8) ÷ 4 = −2 m/s². The minus sign means the acceleration points backwards, opposite to the motion, the same way as the net force (friction). ::: :::keep-together ### Direction **Which way did the net force point while it slowed down?** ☐ Backwards, the same way as the acceleration ☐ There was no net force ☐ Forwards, the way it was moving ::: :::teacher-note **Answer:** Backwards, the same way as the acceleration Acceleration always points the same way as the net force. ::: ::: :::wider ## Elaborate: turning :::keep-together **A car drives around a roundabout at a steady 30 km/h. Is it accelerating?** ☐ Only if it speeds up ☐ No: its speed is constant ☐ Yes: its direction is changing, so its velocity is changing ::: :::teacher-note **Answer:** Yes: its direction is changing, so its velocity is changing There must be a net force towards the centre of the roundabout: the friction of the tyres on the road. ::: ::: ## Evaluate: look back Write one sentence: "At first I thought… Now I think… because…" **Revision summary** - **Acceleration** is the rate of change of velocity: **a = Δv ÷ Δt**, in **m/s²**. - It has a **direction**, the same as the net force. Slowing down is acceleration opposite to the motion (negative). - A fast object at a steady speed has **zero** acceleration. - On a speed–time graph, acceleration is the **gradient**. ## Exit ticket :::keep-together **A sprinter goes from 0 to 9 m/s in 3 s. What is her acceleration?** {{lines-3}} ::: :::teacher-note **Answer:** a = (9 − 0) ÷ 3 = 3 m/s², forwards. :::