A van of mass 2500kg travelling at speed vms⁻¹ experiences a resistance force of kv² N. The constant power of the van's engine is 62.5 kW.
- (a)The steady speed that the van could maintain when moving along a straight horizontal road is 50ms⁻¹. Show that k = 0.5, and find the acceleration of the van when its speed is 25ms⁻¹ on this straight horizontal road.[4]
- (b)The van begins to ascend a hill inclined at an angle θ° to the horizontal. The van travels along a line of greatest slope of the hill. The speed of the van at the start of the hill is 20ms⁻¹, and its acceleration is 5ams⁻². Later, on the same hill, the speed of the van is 30ms⁻¹, and its acceleration is ams⁻². The power of the van's engine remains at 62.5kW, and the resistance force remains at 0.5v² N. Find the value of a and the value of θ.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.
- [M1] For use of Power = DF×v - allow 62.5×103 for 62500, allow 62500=k×50³.
- [A1] k = 0.5 AG-allow a correct equation followed by k=0.5.
- [M1] For N2L with 3 terms; dimensionally correct but allow sign errors. If using 62500/50 (=1250) for the DF then M0.
- [A1] Acceleration = 0.875 m s⁻² Allow 7/8
- [*M1] With 4 relevant terms; allow sign errors; Allow sin/cos mix; condone 30 with 5a, 20 with a, but must be dimensionally correct.
- [A2] A1 for either correct equation.
- [DM1] For attempt to solve for a or θ – from equations with the correct number of relevant terms.
- [A1] θ=3[.00] and a = 0.129 Allow a=31/240 , 0.129167... Allow 0.130 (0.129973...) from using θ=3 but not 0.13 unless greater accuracy seen.
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →