Answered

  1. Measuring motion in the lab.
AA ticker-tape timer marks a tape every $1 / 50\text{ s}$. One tape has evenly spaced dots 6.0 mm apart. Find the velocity of the tape.
BA second tape has dots that get steadily farther apart. What does this say about the motion?
CA 5.0 cm wide card blocks a light gate's beam for 0.020 s. Find the card's speed.
DA strobe photo taken every 0.10 s shows a ball at 0.00, 0.40, 0.80, and 1.20 m. Describe its motion and give its velocity.



Part 3: Uniformly Accelerated Motion (suvat)

List s, u, v, a, t (known and unknown) before choosing an equation, and show units at every step.

  1. Derivations. Starting from the definition of acceleration, $a = (v - u) / t$, complete each step.
ARearrange to get equation (1): $v =$ ____ .
BFor uniform acceleration, $v_{\text{avg}} =$ ____ , so $s =$ ____ (equation 2).
CSubstitute equation (1) into equation (2) to show that $s = ut + \frac{1}{2}at^2$ (equation 3). Show every algebra step.
DUse $t = (v - u) / a$ in equation (2) to show that $v^2 = u^2 + 2as$ (equation 4). Show every algebra step.
  1. For each row, name the suvat equation (1-4) that contains the three known values and the one wanted value.

Verified Answer :

  • Part (a): Velocity is \(\frac{0.006\text{ m}}{0.020\text{ s}} = 0.30\text{ m/s}\).
  • Part (b): Increasing distance between dots indicates acceleration.
  • Part (c): Speed is \(\frac{0.050\text{ m}}{0.020\text{ s}} = 2.5\text{ m/s}\).
  • Part (d): Uniform motion with velocity \(4.0\text{ m/s}\).
  • Derivations provide standard suvat equations: \(v = u+at\), \(s = \frac{u+v}{2}t\), \(s = ut + \frac{1}{2}at^2\), and \(v^2 = u^2 + 2as\).

Explanation

  1. Analysis of Measuring Motion Problems

For part (a), the ticker-tape timer marks a tape every \(\Delta t = \frac{1}{50} = 0.020\) s. The distance between adjacent dots is \(\Delta x = 6.0\) mm, which is equivalent to \(0.006\) m. The velocity \(v\) is defined as the change in displacement per unit time:
\[v = \frac{\Delta x}{\Delta t} = \frac{0.006\text{ m}}{0.020\text{ s}} = 0.30\text{ m/s}.\]
For part (b), if the dots on the ticker tape get steadily farther apart over constant time intervals, it indicates that the object is traveling an increasing distance in each consecutive time interval. This means the velocity of the object is increasing, which corresponds to an accelerated motion.
For part (c), the card has a width of \(w = 5.0\) cm = \(0.050\) m. The time taken to block the light gate is \(\Delta t = 0.020\) s. Assuming the velocity is constant while passing through the gate, the speed is:
\[v = \frac{w}{\Delta t} = \frac{0.050\text{ m}}{0.020\text{ s}} = 2.5\text{ m/s}.\]
For part (d), the strobe photo records positions every \(\Delta t = 0.10\) s at \(0.00\) m, \(0.40\) m, \(0.80\) m, and \(1.20\) m. The displacements between consecutive positions are:
\(\Delta x_1 = 0.40 - 0.00 = 0.40\text{ m}\)
\(\Delta x_2 = 0.80 - 0.40 = 0.40\text{ m}\)
\(\Delta x_3 = 1.20 - 0.80 = 0.40\text{ m}\)
Since the displacement in each equal time interval is constant, the velocity is constant. Calculating the velocity:
\[v = \frac{\Delta x}{\Delta t} = \frac{0.40\text{ m}}{0.10\text{ s}} = 4.0\text{ m/s}.\]
This motion is uniform (constant velocity).

  1. Derivation of Suvat Equations

For part 1(a), starting from the definition of acceleration \(a = \frac{v - u}{t}\), we multiply both sides by \(t\) to get \(at = v - u\), then add \(u\) to both sides to isolate \(v\):
\[v = u + at \quad (\text{Equation 1})\]
For part 1(b), under uniform acceleration, the average velocity is the mean of the initial and final velocities:
\[v_{avg} = \frac{u + v}{2} \quad (\text{Equation 2})\]
The displacement is the product of average velocity and time:
\[s = v_{avg} \cdot t = \left(\frac{u + v}{2}\right)t\]
For part 1(c), substitute Equation 1 (\(v = u + at\)) into Equation 2:
\[s = \left(\frac{u + (u + at)}{2}\right)t\]
\[s = \left(\frac{2u + at}{2}\right)t = \left(u + \frac{1}{2}at\right)t\]
\[s = ut + \frac{1}{2}at^2 \quad (\text{Equation 3})\]
For part 1(d), rearrange Equation 1 to solve for \(t\): \(t = \frac{v - u}{a}\). Substitute this into the expression for displacement:
\[s = \left(\frac{u + v}{2}\right) \left(\frac{v - u}{a}\right)\]
Using the difference of squares identity \((v+u)(v-u) = v^2 - u^2\):
\[s = \frac{v^2 - u^2}{2a}\]
Multiplying both sides by \(2a\):
\[2as = v^2 - u^2\]
\[v^2 = u^2 + 2as \quad (\text{Equation 4})\]
The final answers are:

A\(v = u + at\)
B\(v_{avg} = \frac{u + v}{2}\), \(s = \frac{u + v}{2} \cdot t\)
C\(s = ut + \frac{1}{2}at^2\)
D\(v^2 = u^2 + 2as\)

Examples

In a physics laboratory setting, understanding these motion-measurement techniques is fundamental. For example, a student might use a light gate to measure the speed of a rolling cart on an incline to investigate Newton's second law. By capturing the time it takes for a marker on the cart to pass through the light gate, the student can accurately determine the cart's instantaneous velocity at that point. These calculations form the basis for analyzing mechanical energy or momentum in dynamic systems.

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