1 of 8 0-day streak

Recover motion from acceleration

Question 1 of 8

No calculator

Recover the two motion rules, then interpret the requested result

A point moves along a straight line. Time t is in seconds, signed position s is in metres, velocity v is in m/s and acceleration a is in m/s². The model is valid only for

0\le t\le 4

a(t)=-1

v(0)=5

s(0)=-3

Find the velocity and position at t = 4

(a)1 mark

Which condition fixes each constant in the two anti-differentiation stages?

Use the velocity value for the velocity constant, then the position value for the position constant

The acceleration rule already fixes both constants

Use the position value to fix both constants

Use the velocity value to fix both constants

(b)2 marks

Enter the coefficients of the specific velocity after applying its supplied value, including its constant coefficient

Enter every coefficient in descending powers, including zero for an absent term

v(t)=A_1t+A_0

\(\text{coefficient of }t\) \(\text{constant coefficient}\)

Integrate the specific velocity, then use the supplied position value. Enter every coefficient of the specific position, including its independent constant

Enter every coefficient in descending powers, including zero for an absent term

s(t)=A_2t^{2}+A_1t+A_0

\(\text{coefficient of }t^{2}\) \(\text{coefficient of }t\) \(\text{constant coefficient}\)

(d)1 mark

Select the complete specific position rule satisfying both supplied conditions

s(t)=-\frac{1}{2}t^{2}+5t-3

s(t)=-t+5

s(t)=-\frac{1}{2}t^{2}+5t-2

s(t)=-\frac{1}{2}t^{2}+6t-3

(e)1 mark

Enter the requested velocity, in m/s

Exact value

(f)1 mark

Enter the requested position, in m

Exact value