Notes on Uniform Acceleration and SUVAT (from Transcript)

Key Concepts: Uniform Acceleration and SUVAT

• Uniform (constant) acceleration: a = b4v/b4t, with v changing linearly over time.

• Scalar vs. vector quantities:

  • Speed is scalar (m s^{-1}); Velocity is a vector (m s^{-1}) with direction.

  • Acceleration can be negative depending on the chosen coordinate axis.

• Common one-dimensional motion (SUVAT) equations (constant a):

  • Velocity: $v = u + a t$

  • Displacement: $s = ut + \frac{1}{2} a t^2$

  • Velocity squared: $v^2 = u^2 + 2 a s$

  • Rearranged form for displacement: $s = \frac{v^2 - u^2}{2a}$ (derived from the previous equation)

  • Solve for time (t) when velocity is known: $t = \frac{v - u}{a}$

• Sign conventions and gravity:

  • Gravitational acceleration near Earth's surface: $g \approx 9.8\ \mathrm{m\,s^{-2}}$

  • If upward is taken as positive, gravity is negative: $a = -g$

  • A negative acceleration means either slowing down in the positive direction or speeding up in the negative direction; it indicates direction opposite to the chosen positive axis.

• Graphical interpretation:

  • VelocityaTime (v-t) graph: area under the curve gives displacement $s$.

  • Acceleration
    Time (a-t) graph: area under the curve gives the change in velocity $\Delta v$.

  • For uniformly accelerated motion, the v-t graph is a straight line, the s-t graph is a parabola, and the a-t graph is a horizontal line.

Worked Examples (selected from transcript)

• Worked Example 10.6 (1): Car from rest to 60 km/h (16.67 m s^{-1}) in 8 s

  • Given: $u = 0,\ v = 16.67,\ t = 8\ s$

  • Acceleration: $a = \frac{v-u}{t} = \frac{16.67-0}{8} ≈ 2.08\ \mathrm{m\,s^{-2}}$

  • Displacement: $s = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2}(2.08)(8)^2 ≈ 66.69\ \mathrm{m}$

• Worked Example 10.6 (2): Train from rest travels 30 m in 6 s

  • Given: $u = 0,\ s = 30\ m,\ t = 6\ s$

  • Acceleration: $a = \frac{2s}{t^2} = \frac{2\times 30}{6^2} = \frac{60}{36} ≈ 1.67\ \mathrm{m\,s^{-2}}$

  • Velocity after 6 s: $v = u + a t = 0 + 1.67\times 6 ≈ 10\ \mathrm{m\,s^{-1}}$

• Worked Example 10.7A: Spanner dropped from a sixth-floor window; time to ground = 2.2 s

  • Data: $u = 0,\ t = 2.2\ s,\ g = 9.8\ \mathrm{m\,s^{-2}},\ a = -g = -9.8\ \mathrm{m\,s^{-2}}$

  • Displacement: $s = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2}(-9.8)(2.2)^2 ≈ -23.7\ m$

  • Height: 23.7 m (negative displacement indicates downward direction)

  • Impact velocity: $v = u + a t = 0 + (-9.8)(2.2) ≈ -21.56\ \mathrm{m\,s^{-1}}$

  • Answer notes: speed ≈ 22 m s^{-1} downward; velocity is downward with direction stated.

• Worked Example 10.7A (Alternative framing): Distance and velocity for a drop from a window at rest

  • Directional sign conventions clarified; velocity becomes downward as it accelerates under gravity.

7. A Golfer's Dilemma (vertical drop with horizontal component)

• Given: Golf ball dropped off a cliff with initial horizontal velocity; time to hit ground t = 4.0 s; vertical motion governs height and impact speed.

• Height of cliff:

  • Vertical motion: $s = ut + \tfrac{1}{2} g t^2$ with initial vertical velocity $u = 0$, downward positive or upward positive depending on convention; using downward as positive with g = 9.8 m/s^2 gives $s = \tfrac{1}{2} g t^2 = \tfrac{1}{2} (9.8)(4^2) = 78.4\ m$

• Impact velocity:

  • Vertical component: $v_y = g t = 9.8 \times 4 = 39.2\ \mathrm{m\,s^{-1}}\ (downward)$

  • If there is a horizontal component vx = 20 m/s, then overall impact speed: $v = \sqrt{v</em>x^2 + v_y^2} = \sqrt{20^2 + 39.2^2} ≈ 44.0\ \mathrm{m\,s^{-1}}$

CHECK YOUR LEARNING 10.6 (Key Questions and Answers)

• Q1: Which formula links s, v, t and a? Answer: $s = u t + \frac{1}{2} a t^2$

• Q2: One suvat formula rearranged to give $v^2 = u^2 + 2 a s$; identify the original formula: $v^2 = u^2 + 2 a s$

• Q3: Rearrange $v = u + a t$ to solve for t: $t = \frac{v - u}{a}$

• Q4: What does a negative acceleration mean physically? Answer: It indicates deceleration relative to the chosen positive direction; e.g., gravity acting downward gives a = -g when up is positive.

• Q5: If a man travels from A to B at 30 km/h and must average 60 km/h for the whole trip, what return speed is required?

  • Let outward speed be 30 km/h, total distance 2D. The average speed condition is $\frac{2D}{\tfrac{D}{30} + \tfrac{D}{v}} = 60.$

  • Cancelling D: $\frac{2}{1/30 + 1/v} = 60 \Rightarrow \frac{1}{30} + \frac{1}{v} = \frac{1}{30} \Rightarrow \frac{1}{v} = 0.$

  • Therefore, finite return speed cannot achieve an average of 60 km/h; it would require infinite speed.

• Q6: Acceleration from 60 km/h to 100 km/h in 60 s (straight line):

  • Convert speeds: 60 km/h = 16.67 m/s; 100 km/h = 27.78 m/s.

  • Acceleration: $a = \frac{\Delta v}{\Delta t} = \frac{27.78 - 16.67}{60} ≈ 0.185\ \mathrm{m\,s^{-2}}$

• Q7: Ferrari 0 to 96.5 km/h in 3.98 s:

  • 96.5 km/h = 26.8 m/s; acceleration: $a = \frac{26.8}{3.98} \approx 6.73\ \mathrm{m\,s^{-2}}$

• Q8: Head of rattlesnake acceleration: 50 m s^{-2}; time to reach 27 m s^{-1} from rest if car matched:

  • $t = \frac{v}{a} = \frac{27}{50} = 0.54\ s$

• Q9: Car decelerates from 30 m s^{-1} with a = -1.5 m s^{-2}

  • Time to stop: $t = \frac{v}{|a|} = \frac{30}{1.5} = 20\ s$

  • Distance: $s = ut + \frac{1}{2} a t^2 = 30(20) + \frac{1}{2}(-1.5)(20)^2 = 600 - 300 = 300\ m$

• Q10: ( muon in an electric field ) Note: numbers involve extreme values; conceptually, time to stop from initial speed depends on deceleration $a = \text{given}$ and can be found via $t = \tfrac{\Delta v}{a}$ if applicable.

6.4: CHANGE IN SPEED AND VELOCITY

• Key distinction:

  • Change in speed (scalar): (\Delta\text{speed} = v - u) using magnitudes only.

  • Change in velocity (vector): (\Delta v = v - u) with signs indicating direction.

• Definitions:

  • Average acceleration: $a_{\text{avg}} = \frac{\Delta v}{\Delta t}$

  • Change in velocity: $\Delta v = v - u$

  • Change in speed: $\Delta \text{speed} = |v| - |u|$

• Example 6.4.1: Ball bouncing off a floor

  • Initial velocity: downward (u = -5.0\,\mathrm{m\,s^{-1}})

  • Final velocity after bounce: upward (v = +5.0\,\mathrm{m\,s^{-1}})

  • Change in speed: (\Delta \text{speed} = |v| - |u| = 5 - 5 = 0\,\mathrm{m\,s^{-1}})

  • Change in velocity: (\Delta v = v - u = 5 - (-5) = 10\,\mathrm{m\,s^{-1}}) (upward)

• Example 6.4.2: Contact with floor (golf ball)

  • Initial velocity: (u = -5.0\,\mathrm{m\,s^{-1}}) (downward)

  • Final velocity: (v = +5.0\,\mathrm{m\,s^{-1}}) (upward)

  • Contact time: (\Delta t = 25\ \mathrm{ms} = 0.025\ s)

  • Average acceleration during contact: $a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{(+5) - (-5)}{0.025} = \frac{10}{0.025} = 400\ \mathrm{m\,s^{-2}}$ (upward)

• Try Yourself 6.4.1 (summary): A golf ball hits at 9.0 m s^{-1} downward and rebounds at 7.0 m s^{-1} upward

  • Follow the same steps to find the change in speed and velocity; note the sign conventions when applying vectors.

• 6.4 retrieval and practice:

  • Expressions and signs for average acceleration, velocity changes, and time intervals.

  • Example problem types include: a controlled car decelerating, a squash ball rebounding, a greyhound accelerating, etc.

Practical implications and connections

• Real-world relevance:

  • Vehicle safety: understanding braking distances, peak accelerations, and the limits of human tolerance to acceleration.

  • Sports: golf ball trajectories, bouncing physics, and sprint accelerations.

  • Engineering: design of safety equipment and crash tests relies on SUVAT-type analyses under near-constant acceleration assumptions.

• Limitations:

  • Real motion often involves non-constant acceleration due to air resistance, changing forces, or frictional effects.

  • The simple equations assume one-dimensional motion; multi-dimensional problems require vector treatment and component-by-component analysis.

• Ethical, philosophical, or practical implications:

  • When applying high accelerations (e.g., in crash testing or high-performance vehicles), safety considerations and human tolerance limits are essential.

  • Modeling simplifications should be acknowledged in real-world decision-making (e.g., neglecting air resistance at high speeds may lead to errors).

Quick reference: Common Formulas (LaTeX)

• Velocity-time relation: $v = u + a t$

• Displacement under constant acceleration: $s = u t + \frac{1}{2} a t^2$

• Velocity-squared relation: $v^2 = u^2 + 2 a s$

• Solve for displacement from velocities: $s = \frac{v^2 - u^2}{2a}$

• Time from velocity change: $t = \frac{v - u}{a}$

• Gravitational acceleration: $g \approx 9.8\ \mathrm{m\,s^{-2}}; \ a = -g \text{ if upward is positive}$

Additional Practice (from transcript)

• A motorcycle off a cliff with horizontal velocity: determine height, horizontal landing distance, and impact velocity using vertical motion with constant a = g and horizontal constant velocity.

• A rock thrown horizontally from a cliff: compute time to ground, impact velocity, and horizontal range.

• Practice problems 1-4 and 10 (check your learning 10.6) cover applying the four SUVAT equations, graph interpretations, and sign conventions.

10.7: Applications and Check Your Learning

• Discuss why acceleration is not zero at the top of a vertical launch: even at the peak, gravity provides a downward acceleration; only velocity is zero.

• Determine maximum height for vertical launch: use v^2 = u^2 + 2 a s with v = 0 to solve for s.

• Practice: projectiles, vertical throws, and time-to-ground calculations with gravity as the constant acceleration.

• Conceptual questions explore how accelerations relate to real-world motions, including free-fall, launch, rebound, and deceleration.

Summary

• The transcript reinforces the core SUVAT framework for one-dimensional motion with constant acceleration.

• It blends formula derivations, worked numerical problems, conceptual checks, and practical applications to cars, sports, and everyday motion.

• Key skills: manipulate SUVAT equations, distinguish between speed and velocity, understand sign conventions, interpret areas under graphs, and apply sign-aware vector reasoning to real-world problems.