Date | November 2016 | Marks available | 2 | Reference code | 16N.2.sl.TZ0.9 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
A particle P starts from a point A and moves along a horizontal straight line. Its velocity v cms−1v cms−1 after tt seconds is given by
v(t)={−2t+2,for 0⩽t⩽13√t+4t2−7,for 1⩽t⩽12
The following diagram shows the graph of v.
P is at rest when t=1 and t=p.
When t=q, the acceleration of P is zero.
Find the initial velocity of P.
Find the value of p.
(i) Find the value of q.
(ii) Hence, find the speed of P when t=q.
(i) Find the total distance travelled by P between t=1 and t=p.
(ii) Hence or otherwise, find the displacement of P from A when t=p.
Markscheme
valid attempt to substitute t=0 into the correct function (M1)
eg−2(0)+2
2 A1 N2
[2 marks]
recognizing v=0 when P is at rest (M1)
5.21834
p=5.22 (seconds) A1 N2
[2 marks]
(i) recognizing that a=v′ (M1)
egv′=0, minimum on graph
1.95343
q=1.95 A1 N2
(ii) valid approach to find their minimum (M1)
egv(q), −1.75879, reference to min on graph
1.75879
speed =1.76 (cms−1) A1 N2
[4 marks]
(i) substitution of correct v(t) into distance formula, (A1)
eg∫p1|3√t+4t2−7|dt, |∫3√t+4t2−7dt|
4.45368
distance =4.45 (cm) A1 N2
(ii) displacement from t=1 to t=p (seen anywhere) (A1)
eg−4.45368, ∫p1(3√t+4t2−7)dt
displacement from t=0 to t=1 (A1)
eg∫10(−2t+2)dt, 0.5×1×2, 1
valid approach to find displacement for 0⩽t⩽p M1
eg∫10(−2t+2)dt+∫p1(3√t+4t2−7)dt, ∫10(−2t+2)dt−4.45
−3.45368
displacement =−3.45 (cm) A1 N2
[6 marks]