Cards In This Deck

• $\text{What is the formula for the position vector }\mathbf{r}\text{ of a particle at time }t\text{, given its initial position }{\mathbf{r}}_0\text{ and constant velocity }\mathbf{v}\text{?}$
  $\mathbf{r}={\mathbf{r}}_0+\mathbf{v}t$
• $\text{What is the mathematical condition for two particles }P\text{ and }Q\text{ to collide at time }t\text{?}$
  $$\begin{gathered} \text{Their position vectors must be equal at the same time }t\text{.} \\ {\mathbf{r}}_P={\mathbf{r}}_Q \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{To find the speed of a particle given its position vector }\mathbf{r}=3t\mathbf{i}+4t\mathbf{j}\text{, you calculate }|\mathbf{r}|\text{ and then divide by }t\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False (mostly)} \\ \text{✅ Correct Method: Differentiate }\mathbf{r}\text{ (or identify coefficients of }t\text{ if linear) to find velocity }\mathbf{v}\text{, then find the magnitude }|\mathbf{v}|\text{.} \\ \text{In this specific linear case it works, but conceptually speed is }|\mathbf{v}|\text{, not }|\mathbf{r}|/t\text{ (which is average speed from origin).} \end{gathered}$$
• $\text{Condition for the formulae }v=u+at\text{ and }s=\frac{1}{2}(u+v)t\text{ to be valid?}$
  $\text{The acceleration }a\text{ must be constant (uniform).}$
• $\text{Which SUVAT formula should you select if a problem gives the initial velocity }u\text{, final velocity }v\text{, and acceleration }a\text{, and asks for the displacement }s\text{?}$
  $v^2=u^2+2as$
• $$\begin{gathered} \text{True or False?} \\ \text{If you solve }s=ut+\frac{1}{2}a{t}^2\text{ for time }t\text{ and get }t=2\text{ and }t=-1\text{, both answers are valid in the context of a motion problem starting at }t=0\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: Only }t=2\text{ is valid.} \\ \text{Time }t\text{ represents time elapsed since the start of motion, so }t\text{ cannot be negative (unless the model explicitly considers time before }t=0\text{).} \end{gathered}$$
• $\text{Formula for displacement }s\text{ that eliminates initial velocity }u\text{ (given }v,a,t\text{)?}$
  $s=vt-\frac{1}{2}a{t}^2$
• $$\begin{gathered} \text{True or False?} \\ \text{In the constant acceleration formula }s=ut+\frac{1}{2}a{t}^2\text{, the variable }s\text{ represents the total distance travelled by the particle.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Meaning: }s\text{ represents the displacement from the starting position.} \\ \text{If a particle changes direction (e.g., goes up and comes back down), }s\text{ is the net vector distance from the start, not the total path length covered.} \end{gathered}$$
• $\text{When a particle is dropped from rest, what values should you assign to }u\text{ and }a\text{ (assuming downwards is positive)?}$
  $$\begin{gathered} \text{* }u=0{ms}^{-1} \\ \text{* }a=9.8{ms}^{-2} \end{gathered}$$
• $\text{First step to solve a vertical motion problem where a ball is projected upwards from a point }X\text{ above the ground?}$
  $$\begin{gathered} \text{Define a positive direction (usually upwards or downwards).} \\ \text{If taking upwards as positive:} \\ \text{* }u=+\text{ value} \\ \text{* }a=g=-9.8{ms}^{-2} \\ \text{* Displacement }s\text{ is measured from }X\text{ (so ground level is }s=-h\text{).} \end{gathered}$$
• $\text{What value should you set for }s\text{ to find the time when a particle projected vertically upwards returns to its starting point?}$
  $\text{Set }s=0\text{.}$
• $\text{How do you find the displacement from a velocity-time graph involving both positive and negative velocities?}$
  $$\begin{gathered} \text{Calculate the net area.} \\ (Areaaboveaxis)-(Areabelowaxis) \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{When calculating the area under a velocity-time graph to find the total distance travelled, you simply calculate the net integral (sum of areas) from }t=0\text{ to }t=T\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Method: You must calculate the absolute area of each section.} \\ \text{Areas below the }t\text{-axis (negative velocity) represent distance travelled in the reverse direction. For total distance, add the magnitude of these areas to the positive areas (do not subtract them).} \end{gathered}$$
• $\text{How do you calculate the change in velocity from an acceleration-time graph?}$
  $\text{Calculate the area under the graph (between the line and the time axis) for the specific time interval.}$
• $$\begin{gathered} \text{True or False?} \\ \text{Average speed is calculated using the formula }\frac{u+v}{2}\text{.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: Average Speed = }\frac{TotalDistance}{TotalTime} \\ \frac{u+v}{2}\text{ is only the average velocity and only if acceleration is constant.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{If a car is moving in the negative direction (velocity is negative) and is slowing down (decelerating), its acceleration }a\text{ must be negative.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: The acceleration }a\text{ is positive.} \\ \text{Deceleration means acceleration acts in the opposite direction to motion. If }v\text{ is negative (}-\text{), }a\text{ must be positive (}+\text{) to oppose it.} \end{gathered}$$
• $$\begin{gathered} \text{True or False?} \\ \text{At the greatest height reached by a particle projected vertically upwards, its acceleration is zero.} \end{gathered}$$
  $$\begin{gathered} \text{❌ False} \\ \text{✅ Correct Answer: Acceleration is }g=9.8{ms}^{-2}\text{ (downwards).} \\ \text{At the greatest height, the velocity is zero (}v=0\text{), but gravity is still acting on the object, so acceleration remains constant.} \end{gathered}$$