Date | May 2014 | Marks available | 2 | Reference code | 14M.2.sl.TZ2.9 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
A particle moves in a straight line. Its velocity, v ms−1, at time t seconds, is given by
v=(t2−4)3, for 0⩽
Find the velocity of the particle when t = 1.
Find the value of t for which the particle is at rest.
Find the total distance the particle travels during the first three seconds.
Show that the acceleration of the particle is given by a = 6t{({t^2} - 4)^2}.
Find all possible values of t for which the velocity and acceleration are both positive or both negative.
Markscheme
substituting t = 1 into v (M1)
eg v(1),{\text{ }}{\left( {{1^2} - 4} \right)^3}
velocity = - 27{\text{ }}\left( {{\text{m}}{{\text{s}}^{ - 1}}} \right) A1 N2
[2 marks]
valid reasoning (R1)
eg v = 0,{\text{ }}{\left( {{t^2} - 4} \right)^3} = 0
correct working (A1)
eg {t^2} - 4 = 0,{\text{ }}t = \pm 2, sketch
t = 2 A1 N2
[3 marks]
correct integral expression for distance (A1)
eg \int_0^3 {\left| v \right|,{\text{ }}\int {\left| {{{\left( {{t^2} - 4} \right)}^3}} \right|,{\text{ }} - \int_0^2 {v{\text{d}}t + \int_2^3 {v{\text{d}}t} } } } ,
\int_0^2 {{{\left( {4 - {t^2}} \right)}^3}{\text{d}}t + \int_2^3 {{{\left( {{t^2} - 4} \right)}^3}{\text{d}}t} } (do not accept \int_0^3 {v{\text{d}}t} )
86.2571
{\text{distance}} = 86.3{\text{ (m)}} A2 N3
[3 marks]
evidence of differentiating velocity (M1)
eg v'(t)
a = 3{\left( {{t^2} - 4} \right)^2}(2t) A2
a = 6t{\left( {{t^2} - 4} \right)^2} AG N0
[3 marks]
METHOD 1
valid approach M1
eg graphs of v and a
correct working (A1)
eg areas of same sign indicated on graph
2 < t \leqslant 3 (accept t > 2) A2 N2
METHOD 2
recognizing that a \geqslant 0 (accept a is always positive) (seen anywhere) R1
recognizing that v is positive when t > 2 (seen anywhere) (R1)
2 < t \leqslant 3 (accept t > 2) A2 N2
[4 marks]