Teacher's Guide - Pearson

Project (Student’s Book page 257)1. AB: Constant velocity for 2 minutes, away from A in a positive direction.Displacement is a straight line, which indicates a constant velocity. BC is wherethe object stands still, so the velocity is zero. CD is displacement in a directionopposite to the original direction, therefore it is a negative velocity (since it is inthe opposite direction) but still at a constant velocity, hence the straight line fordisplacement. Note that the time on the velocity-time graph has been convertedto hours in order to express the velocity in km/h.2. The velocity stays constant, i.e. acceleration is zero.3. Area A 5 2 kmArea B 5 26 kmThe areas represent the displacement since km/h 3 h 5 km.4. The displacement is zero in that time interval.5. Yes, the object ends up 4 km from its starting point, but in a negative direction.Activity 14.5 (Student’s Book page 259)1. Velocity 5 4 m/s East after 2 seconds2. Velocity stays constant at 12 m/s East for 14 seconds.3. The scooter stops.4. 0 – 6 seconds5. BC: 20 – 24 seconds6. It accelerates constantly from a standstill for 6 seconds, then travels at a constantvelocity (in a straight line) for 14 seconds, after which it slows down at aconstant rate for 4 seconds until it stops.7. 6 seconds8. Displacement 5 area under velocity-time graph.∴ Displacement 5 ​1__2 ​ 3 4 3 8 5 16 m9. Total area 5 total displacement5 ​1__2 ​ 3 6 3 12 1 14 3 12 1 ​ 1__2 ​ 3 4 3 125 228 metres10. 222 metresChapter 14: Travel graphs83