Date | November 2017 | Marks available | 2 | Reference code | 17N.2.sl.TZ0.9 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Hence or otherwise and Find | Question number | 9 | Adapted from | N/A |
Question
Note: In this question, distance is in metres and time is in seconds.
A particle P moves in a straight line for five seconds. Its acceleration at time tt is given by a=3t2−14t+8a=3t2−14t+8, for 0⩽t⩽5.
When t=0, the velocity of P is 3 ms−1.
Write down the values of t when a=0.
Hence or otherwise, find all possible values of t for which the velocity of P is decreasing.
Find an expression for the velocity of P at time t.
Find the total distance travelled by P when its velocity is increasing.
Markscheme
t=23 (exact), 0.667, t=4 A1A1 N2
[2 marks]
recognizing that v is decreasing when a is negative (M1)
ega<0, 3t2−14t+8⩽0, sketch of a
correct interval A1 N2
eg23<t<4
[2 marks]
valid approach (do not accept a definite integral) (M1)
egv∫a
correct integration (accept missing c) (A1)(A1)(A1)
t3−7t2+8t+c
substituting t=0, v=3 , (must have c) (M1)
eg3=03−7(02)+8(0)+c, c=3
v=t3−7t2+8t+3 A1 N6
[6 marks]
recognizing that v increases outside the interval found in part (b) (M1)
eg0<t<23, 4<t<5, diagram
one correct substitution into distance formula (A1)
eg∫230|v|, ∫54|v|, ∫423|v|, ∫50|v|
one correct pair (A1)
eg3.13580 and 11.0833, 20.9906 and 35.2097
14.2191 A1 N2
d=14.2 (m)
[4 marks]