IB Physics SL A.1 Kinematics questions

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Type

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Question 1

Paper 2Easy9 marks

A cart moves along a straight track in the positive $x$ -direction. Figure 1 shows its velocity $v$ as a function of time $t$ At $t=0$ the cart is at $x=5.0\ \mathrm{m}$

(a)

State the velocity of the cart at $t=4.0\ \mathrm{s}$

[1]

(b)

Determine the acceleration of the cart between $t=2.0\ \mathrm{s}$ and $t=6.0\ \mathrm{s}$

[2]

(c)

Calculate the displacement of the cart between $t=0$ and $t=2.0\ \mathrm{s}$

[1]

(d)

Calculate the displacement of the cart between $t=2.0\ \mathrm{s}$ and $t=6.0\ \mathrm{s}$

[2]

(e)

State the acceleration of the cart between $t=6.0\ \mathrm{s}$ and $t=8.0\ \mathrm{s}$

[1]

(f)

Determine the position of the cart at $t=6.0\ \mathrm{s}$ using your answers to (c) and (d).

[2]

	Displacement components (east, north) / m	Distance travelled / m
A.	(3, 4)	15
B.	(4, 3)	15
C.	(−3, −4)	15
D.	(3, 4)	11

Answer: A. Displacement is final position minus initial position; distance is the length of the route. From S(0,0) to F(3,4), the displacement components are (3−0, 4−0) = (3, 4) m. The route length is 6 + 4 + 4 + 1 = 15 m.

B: Interchanges the east and north components.
C: Reverses the change in position, using start minus finish.
D: Omits the 4 m segment from Q to R when adding the route lengths.
Remember: displacement depends on the initial and final positions; distance includes the whole path.

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