Kinematics 1-D flash cards

1. 1.Distinguish between rest and motion, and why they are relative.

   A body is at rest if its position does not change with time relative to a chosen frame; it is in motion if its position changes with time. Both are relative because they depend entirely on the reference frame chosen — a passenger is at rest w.r.t. the train but in motion w.r.t. the ground.

   Hint: No absolute rest exists; always specify the frame.

2. 2.Define a point object (particle) and when this idealization is valid.

   An object is treated as a point (particle) when its size is negligible compared to the distances involved in its motion, so its internal structure and rotation can be ignored. Example: Earth as a point in its orbit around the Sun.

   Hint: Size << path length.

3. 3.Differentiate distance and displacement.

   Distance is the total path length actually traversed (scalar, always positive, path-dependent). Displacement is the shortest straight-line vector from initial to final position (vector, can be positive, negative or zero, path-independent). Always distance $\ge$ |displacement|.

   Hint: Path length vs change in position.

4. 4.When is the magnitude of displacement equal to the distance travelled?

   Only when the motion is along a straight line in a single (unchanging) direction — no reversal occurs. The moment the body reverses direction, distance starts exceeding |displacement|.

   Hint: Straight line, no turning back.

5. 5.Can displacement be zero while distance is non-zero? Give an example.

   Yes. If a body returns to its starting point, displacement is $0$ but distance equals the total path length. Example: one full lap around a circular track of radius $r$ gives distance $2\pi r$, displacement $0$.

   Hint: Round trip back to start.

6. 6.Define average speed and average velocity.

   Average speed $= \dfrac{\text{total distance}}{\text{total time}}$ (scalar). Average velocity $= \dfrac{\text{displacement}}{\text{total time}} = \dfrac{\Delta x}{\Delta t}$ (vector). Average speed $\ge$ |average velocity|.

   Hint: Distance/time vs displacement/time.

7. 7.Define instantaneous velocity and instantaneous speed.

   Instantaneous velocity $v = \lim_{\Delta t \to 0}\dfrac{\Delta x}{\Delta t} = \dfrac{dx}{dt}$ — the velocity at a specific instant. Instantaneous speed is its magnitude $|v|$. Unlike averages, instantaneous speed always equals |instantaneous velocity|.

   Hint: Limit of average as $\Delta t\to 0$.

8. 8.Why is instantaneous speed always equal to |instantaneous velocity|, but average speed can exceed |average velocity|?

   Over an infinitesimal interval the path is a straight line, so the tiny distance equals |tiny displacement|, forcing equality at an instant. Over a finite interval the path can curve or reverse, making total distance $>$ |displacement|, so average speed can exceed |average velocity|.

   Hint: Infinitesimal path is straight.

9. 9.Define average acceleration and instantaneous acceleration.

   Average acceleration $= \dfrac{\Delta v}{\Delta t} = \dfrac{v-u}{t}$. Instantaneous acceleration $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$. It is the rate of change of velocity; a vector with SI unit $\text{m/s}^2$.

   Hint: Rate of change of velocity.

10. 10.State the SI units and dimensions of speed, acceleration, and jerk.

    Speed: $\text{m/s}$, $[LT^{-1}]$. Acceleration: $\text{m/s}^2$, $[LT^{-2}]$. Jerk (rate of change of acceleration, $da/dt$): $\text{m/s}^3$, $[LT^{-3}]$.

    Hint: Each derivative adds one $T^{-1}$.

11. 11.What does a positive versus negative acceleration tell you about the motion?

    Acceleration sign is relative to the chosen positive axis, not 'speeding up/slowing down'. Speed increases when $v$ and $a$ have the same sign, and decreases (retardation) when $v$ and $a$ have opposite signs.

    Hint: Compare signs of $v$ and $a$, not sign of $a$ alone.

12. 12.Define uniform motion and non-uniform motion.

    Uniform motion: equal displacements in equal intervals of time along a straight line — velocity constant, acceleration zero. Non-uniform motion: unequal displacements in equal intervals — velocity changes, so acceleration is non-zero.

    Hint: Constant velocity vs changing velocity.

13. 13.In uniformly accelerated motion, what stays constant and what changes?

    Acceleration is constant (magnitude and direction). Velocity changes by equal amounts in equal times, while displacement changes by unequal (increasing) amounts. The SUVAT equations apply only in this case.

    Hint: Constant $a$; velocity changes linearly.

14. 14.Write the three primary equations of motion (SUVAT) for constant acceleration.

    $v = u + at$; $\quad s = ut + \tfrac{1}{2}at^2$; $\quad v^2 = u^2 + 2as$. Here $u$ = initial velocity, $v$ = final velocity, $a$ = constant acceleration, $s$ = displacement, $t$ = time.

    Hint: v-t, s-t, and v-s relations.

15. 15.Derive $v = u + at$ using calculus.

    By definition $a = \dfrac{dv}{dt}$, so $dv = a\,dt$. Integrating from $u$ to $v$ and $0$ to $t$ with constant $a$: $\int_u^v dv = a\int_0^t dt \Rightarrow v - u = at \Rightarrow v = u + at$.

    Hint: Integrate $dv = a\,dt$.

16. 16.Derive $s = ut + \tfrac{1}{2}at^2$ using calculus.

    With $v = \dfrac{ds}{dt} = u + at$, we have $ds = (u+at)\,dt$. Integrating: $s = \int_0^t (u+at)\,dt = ut + \tfrac{1}{2}at^2$.

    Hint: Integrate $v = u+at$ over time.

17. 17.Derive $v^2 = u^2 + 2as$ without using time.

    Use $a = v\dfrac{dv}{ds}$ (chain rule), so $v\,dv = a\,ds$. Integrating: $\int_u^v v\,dv = a\int_0^s ds \Rightarrow \tfrac{1}{2}(v^2-u^2) = as \Rightarrow v^2 = u^2 + 2as$.

    Hint: Write $a = v\,dv/ds$.

18. 18.State the formula for displacement in the n-th second and note why it is not a 'per-second' distance.

    $s_{n} = u + \tfrac{a}{2}(2n-1)$. It is the displacement during the single interval between $t=(n-1)$ s and $t=n$ s, obtained as $s_n = s(n) - s(n-1)$; its unit is metre (over that 1 s), not m/s.

    Hint: Difference of two SUVAT displacements.

19. 19.For a body starting from rest with constant acceleration, in what ratio are distances covered in successive equal time intervals?

    Distances in the 1st, 2nd, 3rd,... seconds are in the ratio $1:3:5:7:\dots$ (odd numbers, Galileo's odd-number rule). Total distances after $1,2,3,\dots$ s are in ratio $1:4:9:\dots$ (i.e. $t^2$).

    Hint: $s_n \propto (2n-1)$ from rest.

20. 20.Give the displacement–time graph shapes for (a) rest, (b) uniform velocity, (c) uniform acceleration.

    (a) Rest: horizontal straight line (constant $x$). (b) Uniform velocity: straight line with constant non-zero slope. (c) Uniform acceleration: parabola ($x \propto t^2$ term). Slope of an $x$–$t$ graph = velocity.

    Hint: Slope = velocity.

21. 21.What does the slope of an x–t graph and the slope of a v–t graph represent?

    Slope of $x$–$t$ graph $= \dfrac{dx}{dt} =$ instantaneous velocity. Slope of $v$–$t$ graph $= \dfrac{dv}{dt} =$ instantaneous acceleration.

    Hint: One derivative per graph.

22. 22.What does the area under a v–t graph and under an a–t graph represent?

    Area under $v$–$t$ graph $= \int v\,dt =$ displacement (signed; area below axis is negative). Area under $a$–$t$ graph $= \int a\,dt =$ change in velocity $\Delta v$.

    Hint: Integrate to go up one level.

23. 23.How do you read displacement versus distance from a v–t graph?

    Displacement = algebraic (signed) area, counting area below the time axis as negative. Distance = total unsigned area, treating every region as positive. They differ whenever the velocity changes sign.

    Hint: Signed area vs total |area|.

24. 24.On an x–t graph, how do you identify the instant a body is momentarily at rest or reverses direction?

    The body is momentarily at rest where the tangent (slope) is horizontal, i.e. $v = dx/dt = 0$. It reverses direction where the $x$–$t$ curve reaches a maximum or minimum (turning point).

    Hint: Zero slope = zero velocity.

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