Moderate -0.8 This is a straightforward calculus mechanics question requiring two differentiations and solving a linear equation. The steps are routine: differentiate s to get v, differentiate v to get a, set a=0 to find t, substitute back into v. No problem-solving insight needed, just direct application of standard techniques.
A particle moves in a straight line. The displacement of the particle at time \(t\) s is \(s\) m, where
$$s = t^3 - 6t^2 + 4t.$$
Find the velocity of the particle at the instant when its acceleration is zero. [4]
For using the method of completing the square or using the value
' '
−b
of‘ to find the t value of the minimum velocity
2a
Answer
Marks
M1
Use of the t value at minimum velocity to find v
v=–8ms–1
A1
4
Answer
Marks
Guidance
Question
Answer
Mark
Question 1:
1 | (v=)3t2– 12t + 4 | *M1 | Attempt at differentiation of s to find v
(a=)6t– 12 | *M1 | Attempt at differentiation of v to find a
[When a= 0, t = 2] | DM1 | Solve to find t when a=0 and find v at this time
v=–8ms–1 | A1
Alternative method for question 1
(v=)3t2– 12t + 4 | M1 | Attempt at differentiation of s to find v
(v=)3(t– 2)2 – 8
−b 12
or t= = =2
2a 6 | M1 | For using the method of completing the square or using the value
' '
−b
of‘ to find the t value of the minimum velocity
2a
M1 | Use of the t value at minimum velocity to find v
v=–8ms–1 | A1
4
Question | Answer | Mark | Guidance
A particle moves in a straight line. The displacement of the particle at time $t$ s is $s$ m, where
$$s = t^3 - 6t^2 + 4t.$$
Find the velocity of the particle at the instant when its acceleration is zero. [4]
\hfill \mbox{\textit{CAIE M1 2019 Q1 [4]}}