# A.1 Kinematics SL Paper 1 |IBDP Physics SL 2025 | Exam Style Questions

A.1 Kinematics SL Paper 1 : Exam Style Questions

## IBDP Physics SL Paper 1 – All Chapters

Topic:  A.1 Kinematics SL Paper 1

Syllabus: Displacement, Velocity, Acceleration, Uniform Acceleration Equations, Projectile Motion

#### Questions-A.1 Kinematics SL Paper 1

A block of mass $$2.0 \mathrm{~kg}$$ accelerates uniformly at a rate of $$1.0 \mathrm{~m} \mathrm{~s}^{-2}$$ when a force of $$4.0 \mathrm{~N}$$ acts on it. The force is doubled while resistive forces stay the same. What is the block’s acceleration?

A. $$4.0 \mathrm{~ms}^{-2}$$

B. $$3.0 \mathrm{~m} \mathrm{~s}^{-2}$$

C. $$2.0 \mathrm{~m} \mathrm{~s}^{-2}$$

D. $$1.0 \mathrm{~ms}^{-2}$$

Ans:B

We can use Newton’s second law to find the initial resistive force ($$R_1$$) and the final resistive force ($$R_2$$):

1. For the initial case ($$F_1$$), the net force ($$F_{\text{net}_1}$$) is responsible for the initial acceleration ($$a_1$$):

$F_{\text{net}_1} = F_1 – R_1 = m \cdot a_1$

Solving for $$R_1$$:

$R_1 = F_1 – m \cdot a_1$
$R_1 = 4.0 \, \mathrm{N} – (2.0 \, \mathrm{kg} \cdot 1.0 \, \mathrm{m/s^2}) = 4.0 \, \mathrm{N} – 2.0 \, \mathrm{N} = 2.0 \, \mathrm{N}$

So, in the initial case, the resistive force ($$R_1$$) is $$2.0 \, \mathrm{N}$$.

2. For the final case ($$F_2$$), the net force ($$F_{\text{net}_2}$$) is responsible for the final acceleration ($$a_2$$):

$F_{\text{net}_2} \Rightarrow F_2 – R_1 = m \cdot a_2$

Solving for $$a_2$$:

$2.0 \, \mathrm{kg}\cdot a_2 = 8.0 \, \mathrm{N} – 2.0 \, \mathrm{kg}$

$6.0 \, \mathrm{N} = 2.0 \, \mathrm{kg} \cdot a_2$

$a_2 = 3.0 \, \mathrm{m/s^2}$

#### Question

A car travels clockwise around a circular track of radius R. What is the magnitude of displacement from X to Y?

A. $$R \frac{3 \pi}{2}$$

B. $$R \frac{\pi}{2}$$

C. $$R \sqrt{2}$$

D. $$R$$

Ans:C

The displacement from point X to point Y is a chord of the circle and can be represented by the line connecting points X and Y.

#### Question

A car accelerates uniformly. The car passes point $$X$$ at time $$t_1$$ with velocity $$v_1$$ and point $$Y$$ at time $$t_2$$ with velocity $$v_2$$. The distance $$X Y$$ is $$s$$.

The following expressions are proposed for the magnitude of its acceleration a:

I. $$a=\frac{2 s}{\left(t_2-t_1\right)^2}$$

II. $$a=\frac{v_2^2-v_1^2}{2 s}$$

III. $$a=\frac{v_2-v_1}{t_2-t_1}$$

Which is correct?

A. I and II only

B. I and III only

C. II and III only

D. I, II and III

Ans:C

To determine which expressions for the magnitude of acceleration ($$a$$) are correct, we can analyze each of the proposed expressions one by one:

I. $$a = \frac{2s}{(t_2 – t_1)^2}$$

Expression I is incorrect because it assumes that the initial velocity $$\left(v_1\right)$$ is zero, which may not be the case for the uniformly accelerating car.

II. $$a = \frac{v_2^2 – v_1^2}{2s}$$

This expression is also correct. It’s derived from the kinematic equation for uniformly accelerated motion: $$v^2 = u^2 + 2as$$, where $$v_2$$ is the final velocity, $$v_1$$ is the initial velocity, $$s$$ is the displacement, and $$a$$ is the acceleration. Solving for $$a$$ results in this expression.

III. $$a = \frac{v_2 – v_1}{t_2 – t_1}$$

This expression is also correct and is derived from the definition of acceleration as the rate of change of velocity over time.

So,  expressions (II, and III) are correct.

#### Question

P and Q leave the same point, travelling in the same direction. The graphs show the variation with time t of velocity v for both P and Q.

What is the distance between P and Q when t = 8.0 s?

A 20 m

B 40 m

C 60 m

D 120 m

Ans: B

Distance travelled by P$$=\frac{1}{2}(2)\times 10+(5\times 20)$$
$$=20+100\rightarrow 120$$
Distance travelled by $$Q=\frac{1}{2}(8)\times 40$$
$$=160$$
so , the difference $$= 160- 120 \rightarrow 40$$

#### Question

The velocity–time graph for an accelerating object that is traveling in a straight line is shown below.

Which of the following is the change in displacement of the object in the first 5.0 seconds?

A. 25.0 m
B. 12.5 m
C. 5.0 m
D. 1.0 m

$$Area = \frac{1}{2}\times 5\times 5$$
$$= 12.5m$$