Date | November 2019 | Marks available | 4 | Reference code | 19N.2.SL.TZ0.S_10 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | S_10 | Adapted from | N/A |
Question
A rocket is travelling in a straight line, with an initial velocity of 140 m s−1. It accelerates to a new velocity of 500 m s−1 in two stages.
During the first stage its acceleration, a m s−2, after t seconds is given by a(t)=240sin(2t), where 0⩽t⩽k.
The first stage continues for k seconds until the velocity of the rocket reaches 375 m s−1.
Find an expression for the velocity, v m s−1, of the rocket during the first stage.
Find the distance that the rocket travels during the first stage.
During the second stage, the rocket accelerates at a constant rate. The distance which the rocket travels during the second stage is the same as the distance it travels during the first stage.
Find the total time taken for the two stages.
Markscheme
recognizing that v=∫a (M1)
correct integration A1
eg −120cos(2t)+c
attempt to find c using their v(t) (M1)
eg −120cos(0)+c=140
v(t)=−120cos(2t)+260 A1 N3
[4 marks]
evidence of valid approach to find time taken in first stage (M1)
eg graph, −120cos(2t)+260=375
k=1.42595 A1
attempt to substitute their v and/or their limits into distance formula (M1)
eg ∫1.425950|v|, ∫260−120cos(2t), ∫k0(260−120cos(2t))dt
353.608
distance is 354 (m) A1 N3
[4 marks]
recognizing velocity of second stage is linear (seen anywhere) R1
eg graph, s=12h(a+b), v=mt+c
valid approach (M1)
eg ∫v=353.608
correct equation (A1)
eg 12h(375+500)=353.608
time for stage two =0.808248 (0.809142 from 3 sf) A2
2.23420 (2.23914 from 3 sf)
2.23 seconds (2.24 from 3 sf) A1 N3
[6 marks]