1a Figure 1 General shape of the graph is correct. i.e. horizontal line, followed by negative gradient, followed by a positive gradient. Vertical axis labelled correctly. Horizontal axis labelled correctly. M1 3.3 4th Use and interpret graphs of velocity against time. 1b Makes an attempt to find the area of trapezoidal section where T the car is decelerating. For example, 15 10 is seen. 4 Makes an attempt to find the area of the trapezoidal section 3T where the car is accelerating. For example, 10 0 is 4 seen. 5T 90T States that 15T 131.5 4 4 (3) 4th Calculate and interpret areas under velocity time graphs. Solves to find the value of T: T = 30 (s). (7 marks) 1a Accept the horizontal axis labelled with the correct intervals. 1b Award full marks for correct final answer, even if some work is missing. Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.
a Velocity = acceleration time seen or implied. M1 3.1b 4th Velocity = 11 8 = 88 m s 1 Figure General shape of the graph is correct. i.e. positive gradient, followed by horizontal line, followed by negative gradient not returning to zero. M1 3.3 Use and interpret graphs of velocity against time. Vertical axis labelled correctly. Horizontal axis labelled correctly. b Makes an attempt to find the area of the trapezoidal section. 1 For example, 88 40 is seen. Demonstrates an understanding that the three areas must total 1404. For example, 1 888 88T 1 88 40 1404 or 35 88T 18 1404 is seen. (5) 4th M1.1 Calculate and interpret areas under velocity time graphs. Correctly solves to find T = 10.5 (s). (3) (8 marks) a Accept the horizontal axis labelled with the correct intervals. b Award full marks for correct final answer, even if some work is missing. Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.
3a v u a seen or implied. t 0 6 14 Finds the value of a: a 0.4 m s 35 35 M1 3.1b 5th () 3b t Use the fact that 1 t equivalent. 4 to write 3t 1 = 4t or 3t 1 4t = 0 or 3 States or implies that t 1 + t = 35 M1 3.1b 5th Solves to find t 1 = 0 or t = 15. Could use substitution or simultaneous equations. Does not need to find both values for mark to be awarded as either value can be used going forward. Use v = u + at to write either x = 6 + 0.4(0) or 0 = x + 0.4(15) M1.a Finds x = 14 (m s 1 ). A1ft 1.1b (5) 3c u v States or implies that s t 6 0 35 455 (m). Finds the value of s: s M1.a 5th () (9 marks) 3b Award ft marks for a correct answer using their value from part a. Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.
4a Demonstrates an understanding of the need to use 1 s ut at This can implied by using the equation in the next step(s). Demonstrates the need to use (t 3) when finding the displacement of Q from A (or use (t + 3) when finding the displacement of P from A). Can be implied in either of the following steps. Displacement of P: M1 3.1b 5th M1 3.1b.8 0.06 s t t Displacement of Q: 4b Writes s.4 t 3 0.1 t 3.8t 0.06t.4 t 3 0.1 t 3 M1 3.1b 5th Makes an attempt to simplify this equation. For example,.8t 0.06t.4t 7. 0.1 t 6t 9.8t 0.06t.4t 7. 0.1t 0.6t 0.9 0.04t t 6.3 0 Simplifies this expression to t 50t 315 0 4c Makes an attempt to use the quadratic formula: 50 50 4 315 t Solves to find t = 30.1... (s). Could also show that t 5.1... (s). States or implies 1 s ut at Makes a substitution using their 30.1 into the formula: s.830... 1 0.130... (3) M1.a 5th M1 3.1b Finds s = 139.36... (m). Accept awrt 139 (m). A1 ft 1.1b (5) (1 marks) Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.
4a Award both accuracy marks if the following is seen: Displacement of P from A: s.8 t 3 0.06 t 3 Displacement of Q from A: s.4t 0.1t 4c Award ft marks for a correct answer using their 30.. They will have previously lost the first accuracy mark. 5a States or implies that s = 80 B1 3.1b 5th States or implies that a = 9.8 B1 3.1b problems Writes v u as or makes a substitution M1 3.1b involving vertical motion. v 16 9.8 80 Finds v = 4.70... (m s 1 ). Accept awrt 4.7 (m s 1 ). 5b States or implies that s = 5 m. B1 3.1b 5th Simplifies 5 16t 4.9t to obtain 4.9t 16t 5 0 problems Makes an attempt to use the quadratic formula: involving vertical motion. 4.9 16 16 4 4.9 5 t Solves to find t = 0.35 (s). Accept awrt 0.35 (s). Solves to find t =.91 (s). Accept awrt.9 (s). States that the ball is above 85 m for.56 (s). Accept awrt.6 (s). B1 3.a (6) Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.
5c 5c States or implies that at the greatest height v = 0 B1 3.1b 5th 1 M1 3.1b Finds the value of u: u 4.7... 8.54... (m s 1 ). Accept 5 problems awrt 8.5 (m s 1 ). Writes v u as or makes a substitution M1 3.1b involving vertical motion. 0 8.54... 9.8s Finds s = 3.7...(m). Accept awrt 3.7 (m). A1 ft 1.1b Award ft marks for a correct answer using their answer from part a. (14 marks) Education Ltd 017. Copying permitted for purchasing institution only. This material is not copyright free.