Mathematical Methods — Question 10

QCAA 2024, Paper 2 · 1 mark

Q10 · 2024 · Technology-activeSimple familiar1 mark

QUESTION 10

The velocity (m s

^{-1}

) at time

t

(s) of an object is given by

v(t)=0.4t^2+3t

for

t\ge0

. The change in displacement (m) of the object from four to five seconds is

(A)

15.43

(B)

21.63

(C)

32.53

(D)

54.17

ABCD

WORKED SOLUTION

1 mark

ANSWER

B — 21.63

Worked solution

\Delta s=\int_4^5(0.4t^2+3t)\,dt=[(2/15)t^3+(3/2)t^2]_4^5\approx21.63

m, so option B.

Selects option B.

[1 mark]

Worked explanation by QCE Vault; correct option verified against the QCAA answer key.