Understand Kinematics - IB Physics

1. A stone is thrown vertically upwards from the ground. It reaches a maximum height $H$ and falls back to the ground. Air resistance is negligible. Taking the upward direction as positive, which statement is correct regarding the signs of displacement, velocity, and acceleration while the stone is in the air?
   Answer: The acceleration is always negative, while the velocity is first positive then negative. Acceleration due to gravity is always directed downwards, so it is always negative in this coordinate system. The stone's velocity is initially positive (upwards), becomes zero at the maximum height, and is negative (downwards) on its return trip. Therefore, statement A is correct. Distractor B is wrong because velocity becomes negative. Distractor C is wrong because on the way down, both velocity and acceleration are negative, making their product positive. Distractor D is wrong because acceleration is constant and always negative.
2. A train starts from rest and accelerates at $a_1$ for a time $t_1$. It then immediately decelerates at $a_2$ until it comes to rest. What is the total distance travelled by the train?
   Answer: $\frac{1}{2} a_1 t_1^2 (1 + \frac{a_1}{a_2})$ Stage 1 (acceleration): Distance $s_1 = \frac{1}{2}a_1 t_1^2$. The maximum velocity reached is $v_{max} = a_1 t_1$. Stage 2 (deceleration): The train decelerates from $v_{max}$ to 0. Let the time taken be $t_2$. $0 = v_{max} - a_2 t_2 \Rightarrow t_2 = \frac{v_{max}}{a_2} = \frac{a_1 t_1}{a_2}$. The distance covered is $s_2 = v_{max}t_2 - \frac{1}{2}a_2 t_2^2 = (a_1 t_1)(\frac{a_1 t_1}{a_2}) - \frac{1}{2}a_2(\frac{a_1 t_1}{a_2})^2 = \frac{a_1^2 t_1^2}{a_2} - \frac{1}{2}\frac{a_1^2 t_1^2}{a_2} = \frac{1}{2}\frac{a_1^2 t_1^2}{a_2}$. Total distance $s = s_1 + s_2 = \frac{1}{2}a_1 t_1^2 + \frac{1}{2}\frac{a_1^2 t_1^2}{a_2} = \frac{1}{2}a_1 t_1^2 (1 + \frac{a_1}{a_2})$. This is a multi-step problem requiring derivation, making it challenging. The distractors represent plausible but incorrect algebraic combinations.
3. Two stones, P and Q, are thrown from the same point at the same time with the same initial speed. Stone P is thrown vertically upwards, and stone Q is thrown at 30° to the horizontal. Neglecting air resistance, what is true about the stones just after launch?
   Answer: They have the same initial acceleration. Once the stones are in the air, the only force acting on both of them is gravity (neglecting air resistance). Therefore, both stones have the same constant downward acceleration, $g$. This is true for any projectile motion near the Earth's surface. Distractor B is incorrect; P has initial vertical velocity $u$, while Q has $u \sin(30°) = 0.5u$. Distractor C is incorrect; since their initial vertical velocities are different, they will reach different maximum heights. Distractor D is incorrect; P has zero horizontal velocity, while Q has a non-zero horizontal velocity.
4. A particle moves along a straight line with velocity given by $v(t) = 6t^2 - 18t + 12$ m/s, where t is in seconds. At what time does the particle momentarily come to rest for the second time?
   Answer: t = 2.0 s, when the particle changes direction from negative back to positive velocity The particle comes to rest when $v(t) = 0$: $6t^2 - 18t + 12 = 0$. Dividing by 6: $t^2 - 3t + 2 = 0$, which factors as $(t-1)(t-2) = 0$. So the particle stops at t = 1 s and t = 2 s. The second time is t = 2 s. Choice A gives the first time. Choice C refers to maximum displacement, which occurs at a different time. Choice D is incorrect since we found two rest times.
5. Two particles start at the same point and travel in the same direction. Particle A has constant velocity 8 m/s. Particle B starts from rest with constant acceleration 1.6 m/s². After how much time will particle B be 10 m ahead of particle A?
   Answer: t = 12.5 s, calculated by solving the position difference equation Position equations: $x_A = 8t$ and $x_B = \frac{1}{2}(1.6)t^2 = 0.8t^2$. For B to be 10 m ahead: $x_B - x_A = 10$, so $0.8t^2 - 8t = 10$, giving $0.8t^2 - 8t - 10 = 0$. Multiplying by 1.25: $t^2 - 10t - 12.5 = 0$. Using the quadratic formula: $t = \frac{10 ± \sqrt{100 + 50}}{2} = \frac{10 ± \sqrt{150}}{2} = \frac{10 ± 5\sqrt{6}}{2}$. Taking the positive root: $t ≈ 12.25$ s ≈ 12.5 s. Choice A is when velocities are equal ($v_B = v_A$: $1.6t = 8$, so $t = 5$ s). Choice B is when they meet ($x_A = x_B$).
6. An object's position $x$ as a function of time $t$ is given by $x = At^3 - Bt$, where A and B are positive constants. At what time is the object's acceleration equal to zero?
   Answer: 0 To find acceleration, we must differentiate the position function twice with respect to time. Velocity $v = \frac{dx}{dt} = 3At^2 - B$. Acceleration $a = \frac{dv}{dt} = 6At$. We want to find the time $t$ when $a=0$. Setting $6At = 0$, and since $A$ is a positive constant, we find that $t=0$. Distractor C gives the time when the velocity is zero. Distractor B is an algebraic error. Distractor D incorrectly assumes the acceleration is constant.
7. A car travelling at speed $v$ can brake to a stop in a distance $d$. Assuming the same constant braking force, what is the stopping distance if the car is initially travelling at speed $3v$?
   Answer: $9d$ We can use the kinematic equation $v_f^2 = v_i^2 + 2as$. Here, the final speed $v_f = 0$, initial speed is $v_i$, acceleration is $a$ (which is negative), and stopping distance is $s=d$. So, $0 = v_i^2 + 2ad \Rightarrow d = \frac{-v_i^2}{2a}$. This shows that the stopping distance $d$ is proportional to the square of the initial speed ($d \propto v_i^2$), since the acceleration $a$ is constant (due to constant braking force). If the initial speed is tripled to $3v_i$, the new stopping distance $d'$ will be $d' \propto (3v_i)^2 = 9v_i^2$. Therefore, the new stopping distance is $9d$. Distractor B is a common error from assuming a linear relationship between speed and stopping distance.
8. A small sphere is released from rest in a tall column of viscous liquid. Which statement correctly describes the net force on the sphere as it falls?
   Answer: The net force is initially at its maximum value and decreases, approaching zero. The net force on the sphere is $F_{net} = F_{gravity} - F_{buoyancy} - F_{drag}$. At $t=0$, the sphere is at rest, so the drag force (which depends on speed) is zero. The net force is $F_{gravity} - F_{buoyancy}$, which is its maximum value. As the sphere accelerates and its speed increases, the drag force increases, opposing the motion. This causes the net downward force to decrease. If the column is tall enough, the sphere will reach terminal velocity, where the drag force becomes large enough that the net force approaches zero. Distractor A has the logic reversed. Distractor B would imply constant acceleration, which is false. Distractor D would imply no motion.
9. A boat, which has a speed of $v$ in still water, is to cross a river of width $W$. The river flows at a constant speed $u$. To cross the river in the shortest possible time, the boat should be steered in which direction relative to the river banks?
   Answer: Directly perpendicular to the river banks. The time to cross the river is determined by the component of the boat's velocity that is perpendicular to the river flow. Time $t = \frac{W}{v_{\perp}}$. To minimize the time $t$, the perpendicular component of the velocity, $v_{\perp}$, must be maximized. The boat's velocity relative to the water is $v$. This component is maximized when the boat is pointed directly across the river, perpendicular to the banks, making $v_{\perp} = v$. Distractor B describes the path to minimize the distance travelled downstream (to go straight across), not the time taken.
10. A projectile is launched with a certain speed $u$. For a given horizontal range $R$, which is less than the maximum possible range, it is known that two launch angles, $\theta_1$ and $\theta_2$, will result in hitting the target. What is the relationship between these two angles?
    Answer: $\theta_1 + \theta_2 = 90°$ The range equation for a projectile is $R = \frac{u^2 \sin(2\theta)}{g}$. For a fixed $u$ and $R$, we need $\sin(2\theta_1) = \sin(2\theta_2)$. The sine function has the property that $\sin(x) = \sin(180° - x)$. Therefore, if $2\theta_1 \neq 2\theta_2$, we must have $2\theta_1 = 180° - 2\theta_2$. Dividing by 2 gives $\theta_1 = 90° - \theta_2$, or $\theta_1 + \theta_2 = 90°$. The two angles are complementary. Distractor C is only true for the maximum range, where $\theta_1 = \theta_2 = 45°$. The other distractors represent common misconceptions.
11. An elastic ball is dropped from rest onto a hard horizontal surface and bounces once. The effects of air resistance and the duration of contact with the surface are negligible. Taking the upward direction as positive, which statement best describes the acceleration-time graph of the motion?
    Answer: A constant horizontal line at a negative value. While the ball is in the air (both falling and rising), the only force acting on it is gravity (since air resistance is negligible). Therefore, its acceleration is constant and equal to $-g$. The question states the duration of contact is negligible, implying we consider only the time the ball is in free-fall. During this entire time, the acceleration is constant. The large upward acceleration during the bounce itself occurs over a negligible time and would not appear on a graph representing the overall motion. Distractor B correctly identifies the acceleration during contact, but the question focuses on the motion where contact time is negligible, so the dominant feature is constant acceleration due to gravity. Distractor C has the wrong sign. Distractor D describes a changing acceleration.
12. An object moves with a constant, non-zero acceleration. Which statement about the object's motion must be true?
    Answer: Its velocity changes at a constant rate. The definition of constant acceleration is that the velocity changes by the same amount in each unit of time, i.e., it changes at a constant rate. Distractor A is false; if acceleration is opposite to the initial velocity, the object will slow down (speed decreases). Distractor C is false; in projectile motion, the acceleration is constant (due to gravity), but the direction of motion continuously changes. Distractor D is false; for accelerated motion, the distance covered in successive time intervals increases (if speeding up) or decreases (if slowing down), as shown by the $t^2$ term in $s = ut + \frac{1}{2}at^2$.
13. A stone is thrown from the top of a 45 m high cliff with a horizontal velocity of 10 m s⁻¹. What is the speed of the stone just before it strikes the ground? (Use $g = 10$ m s⁻²).
    Answer: 32 m s⁻¹ The horizontal component of the velocity, $v_x$, remains constant at 10 m s⁻¹. The vertical component of velocity, $v_y$, can be found using $v_y^2 = u_y^2 + 2as$. Here, initial vertical velocity $u_y = 0$, $a = g = 10$ m s⁻², and displacement $s = 45$ m. So, $v_y^2 = 0^2 + 2(10)(45) = 900$, which gives $v_y = 30$ m s⁻¹. The final speed is the magnitude of the resultant velocity vector: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 30^2} = \sqrt{100 + 900} = \sqrt{1000}$. $\sqrt{1000} \approx 31.6$ m s⁻¹, which is closest to 32 m s⁻¹. Distractor B is only the final vertical component of velocity. Distractor D is the arithmetic sum of the components, which is incorrect for vector addition.
14. A ball is thrown vertically upwards and returns to its starting point. Air resistance is significant. How does the time taken to rise to its maximum height, $t_{up}$, compare with the time taken to fall back, $t_{down}$?
    Answer: $t_{up} < t_{down}$ On the way up, both gravity and air resistance act downwards. The net downward acceleration $a_{up}$ is greater than $g$. On the way down, gravity acts downwards but air resistance acts upwards, so the net downward acceleration $a_{down}$ is less than $g$. Since the ball travels the same vertical distance up and down, but the average magnitude of acceleration is greater on the way up, the time taken for the upward journey must be shorter. $t_{up} < t_{down}$. Distractor B is only true in a vacuum. Distractor C has the reasoning reversed. Distractor D is incorrect because while the values of the times depend on the speed, the inequality $t_{up} < t_{down}$ holds regardless of the initial speed.
15. A stone is thrown vertically upward with initial velocity 15 m/s from the top of a 20 m tall cliff. What is the velocity of the stone when it passes the initial launching point on its way down?
    Answer: -15 m/s, since the stone returns to its starting height with the same speed but opposite direction When analyzing projectile motion problems involving a return to the starting position, you're dealing with a fundamental principle of energy conservation and kinematic symmetry. The key insight is that when an object returns to its original height, it has the same kinetic energy (and thus the same speed) as when it started. Since the stone begins with an upward velocity of 15 m/s and returns to the exact same height, it must have a speed of 15 m/s when it passes the launching point again. However, since it's now moving downward, the velocity is -15 m/s. You can verify this using kinematics: $v^2 = v_0^2 + 2a\Delta y$. When the stone returns to its starting point, $\Delta y = 0$, so $v^2 = v_0^2 = (15)^2$, giving $v = -15$ m/s (negative because it's moving downward). Option A incorrectly suggests using "average acceleration," which is meaningless here since gravity provides constant acceleration. Option B arbitrarily adds extra acceleration that doesn't exist - gravity acts consistently at 9.8 m/s² throughout the motion. Option C appears to use some incorrect "net effect" calculation that has no basis in physics principles. The correct answer is D: -15 m/s, demonstrating that projectiles return to their starting height with the same speed but opposite direction. Study tip: For any projectile motion problem where an object returns to its starting height, the final speed always equals the initial speed - this symmetry principle can save you calculation time on exams.
16. A car accelerates uniformly from rest. It travels 40 m in the third second of its motion. What is its acceleration? The first second is from t=0 to t=1 s.
    Answer: 16 m s⁻² The third second of motion is the interval between $t=2$ s and $t=3$ s. The distance travelled in time $t$ is $s(t) = ut + \frac{1}{2}at^2$. Since $u=0$, $s(t) = \frac{1}{2}at^2$. Distance at $t=3$ s: $s(3) = \frac{1}{2}a(3^2) = 4.5a$. Distance at $t=2$ s: $s(2) = \frac{1}{2}a(2^2) = 2a$. The distance travelled in the third second is $s(3) - s(2) = 4.5a - 2a = 2.5a$. We are given this distance is 40 m. So, $2.5a = 40 \Rightarrow a = \frac{40}{2.5} = 16$ m s⁻². Distractors arise from incorrect interpretation of 'third second' or calculation errors, e.g., setting $s(3) = 40$, which gives $a \approx 8.9$, or confusing average velocity.
17. A projectile is launched at an angle of 30° above horizontal with initial speed 20 m/s from ground level. What is the speed of the projectile when it returns to ground level?
    Answer: 20 m/s, since mechanical energy is conserved and the projectile returns to its initial height By conservation of mechanical energy, when the projectile returns to the same height (ground level), it must have the same kinetic energy and therefore the same speed as initially: 20 m/s. Choice A gives only $v_x = 20\cos(30°) = 17.3$ m/s. Choice B incorrectly gives the horizontal component as the final speed. Choice D incorrectly adds velocity components: $17.3 + 10 = 27.3$ m/s (approximately 28.3), but velocities add vectorially, not algebraically.
18. A train traveling at 25 m/s applies its brakes and comes to a stop in 250 m. A car traveling at 15 m/s applies brakes with the same deceleration. What distance does the car require to stop?
    Answer: 90 m, calculated using the proportional relationship between stopping distance and initial velocity For the train: $v^2 = v_0^2 + 2as$ gives $0 = (25)^2 + 2a(250)$, so $a = -\frac{625}{500} = -1.25$ m/s². For the car with same deceleration: $0 = (15)^2 + 2(-1.25)s$, so $s = \frac{225}{2.5} = 90$ m. Choice B assumes constant braking time, but stopping distance depends on $v^2$, not $v$. Choice C incorrectly involves mass. Choice D arbitrarily assumes different braking efficiency.
19. A golf ball is hit at an angle above the horizontal. Air resistance has a significant effect on the flight of the ball. Which statement best describes the trajectory of the ball compared to the ideal parabolic path in a vacuum?
    Answer: The maximum height achieved is lower and the horizontal range is shorter. Air resistance is a dissipative force that removes energy from the system. This means the ball will not travel as high or as far as it would in a vacuum. Therefore, both the maximum height and the horizontal range are reduced. Distractor B is incorrect because air resistance makes the trajectory asymmetric; the descent is typically steeper than the ascent. Distractor C is incorrect; because the maximum height is lower and the ball is slowed in both horizontal and vertical directions, the time of flight is shorter. Distractor D is incorrect; the descent is steeper, meaning the angle of impact is greater than the launch angle.
20. A projectile is launched from ground level with kinetic energy $E$ at an angle of 60° to the horizontal. Neglecting air resistance, what is the kinetic energy of the projectile at the highest point of its trajectory?
    Answer: $0.25 E$ The initial kinetic energy is $E = \frac{1}{2}mu^2$, where $u$ is the initial speed. At the highest point of the trajectory, the vertical component of velocity is zero, but the horizontal component remains constant: $v_x = u \cos\theta$. The speed at the highest point is $v = u \cos(60°)$. Since $\cos(60°) = 0.5$, the speed is $v = 0.5u$. The kinetic energy at this point is $E_{top} = \frac{1}{2}mv^2 = \frac{1}{2}m(0.5u)^2 = \frac{1}{2}mu^2(0.5)^2 = E \times 0.25$. Distractor A incorrectly assumes the speed is zero at the top. Distractor C forgets to square the cosine term. Distractor D uses $\sin^2(60°)$, which relates to the potential energy gained.

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