Physics worksheet featuring kinematics practice problems focused on uniform accelerated motion and the four core equations.

Let's solve each of these kinematics problems step by step, using the provided equations and applying the 7-step method where applicable (identify knowns, unknowns, choose equation, plug in values, solve, check units, interpret answer).

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Part 1: Uniform Accelerated Motion

We'll use the following kinematic equations:

- $ a_{\text{ave}} = \frac{\Delta v}{\Delta t} $
- $ d_f = d_i + v_i t + \frac{1}{2} a_{\text{ave}} t^2 $
- $ v_f = v_i + a_{\text{ave}} t $
- $ v_f^2 = v_i^2 + 2a_{\text{ave}}(d_f - d_i) $

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#### Problem 1:
> A rock is falling downwards at 12.0 m/s. After 2.0 seconds it is falling at 31.6 m/s. What was its acceleration?

Step 1: Identify knowns and unknowns

- $ v_i = 12.0 \, \text{m/s} $ (initial velocity)
- $ v_f = 31.6 \, \text{m/s} $ (final velocity)
- $ t = 2.0 \, \text{s} $
- $ a = ? $

Step 2: Choose appropriate equation

Use:
$$
v_f = v_i + a t
$$

Step 3: Plug in values
$$
31.6 = 12.0 + a(2.0)
$$

Step 4: Solve
$$
31.6 - 12.0 = 2.0a \\
19.6 = 2.0a \\
a = \frac{19.6}{2.0} = 9.8 \, \text{m/s}^2
$$

✔ This matches gravity! So, acceleration = 9.8 m/s² downward.

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#### Problem 2:
> An NFL caliber wide receiver can reach a top speed of 10.0 m/s in only 2.4 s. What is their acceleration during this time?

Assume starting from rest: $ v_i = 0 $

- $ v_f = 10.0 \, \text{m/s} $
- $ t = 2.4 \, \text{s} $
- $ a = ? $

Use:
$$
v_f = v_i + a t
$$
$$
10.0 = 0 + a(2.4)
$$
$$
a = \frac{10.0}{2.4} = 4.17 \, \text{m/s}^2
$$

✔ Acceleration = 4.17 m/s²

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#### Problem 3:
> A student is gliding along on a scooter at 2.8 m/s when Mr. Jones walks around the corner and the two collide. If the student is brought to rest in 0.15 s, what is their acceleration? (Hint: should it be acceleration or deceleration?)

- $ v_i = 2.8 \, \text{m/s} $
- $ v_f = 0 \, \text{m/s} $ (brought to rest)
- $ t = 0.15 \, \text{s} $
- $ a = ? $

Use:
$$
v_f = v_i + a t
$$
$$
0 = 2.8 + a(0.15)
$$
$$
-2.8 = 0.15a \\
a = \frac{-2.8}{0.15} = -18.67 \, \text{m/s}^2
$$

✔ The negative sign indicates deceleration (slowing down).
So, acceleration = -18.7 m/s² (or deceleration of 18.7 m/s²)

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#### Problem 4:
> A car is moving at 30.0 km/h when it accelerates at 2.0 m/s² for 3.6 s. What is the car's final speed?

First, convert initial speed to m/s:
$$
30.0 \, \text{km/h} = \frac{30.0 \times 1000}{3600} = 8.33 \, \text{m/s}
$$

Now:
- $ v_i = 8.33 \, \text{m/s} $
- $ a = 2.0 \, \text{m/s}^2 $
- $ t = 3.6 \, \text{s} $
- $ v_f = ? $

Use:
$$
v_f = v_i + a t
$$
$$
v_f = 8.33 + (2.0)(3.6) = 8.33 + 7.2 = 15.53 \, \text{m/s}
$$

Convert back to km/h if needed:
$$
15.53 \, \text{m/s} \times \frac{3600}{1000} = 55.9 \, \text{km/h}
$$

✔ Final speed = 15.5 m/s or 55.9 km/h

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Part 2: Using the 4 Equations and 7-Step Method

#### Problem 1:
> An airplane accelerates down a runway at 5.20 m/s² for 32.8 s until it lifts off. Determine the distance traveled before takeoff.

We are given:
- $ a = 5.20 \, \text{m/s}^2 $
- $ t = 32.8 \, \text{s} $
- Assume starts from rest → $ v_i = 0 $
- $ d = ? $

Use:
$$
d_f = d_i + v_i t + \frac{1}{2} a t^2
$$
Since $ d_i = 0 $, $ v_i = 0 $:
$$
d = \frac{1}{2} a t^2
$$

Plug in:
$$
d = \frac{1}{2}(5.20)(32.8)^2
$$
First compute $ 32.8^2 = 1075.84 $

Then:
$$
d = 0.5 \times 5.20 \times 1075.84 = 2.60 \times 1075.84 = 2800.184 \, \text{m}
$$

✔ Distance ≈ 2800 meters (or 2.80 × 10³ m)

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✔ Final Answers Summary:

#### Part 1:
1. Acceleration of rock: $ \boxed{9.8 \, \text{m/s}^2} $
2. Receiver’s acceleration: $ \boxed{4.17 \, \text{m/s}^2} $
3. Student’s acceleration: $ \boxed{-18.7 \, \text{m/s}^2} $ (deceleration)
4. Car’s final speed: $ \boxed{15.5 \, \text{m/s}} $ or $ \boxed{55.9 \, \text{km/h}} $

#### Part 2:
1. Distance traveled by airplane:
$$
d = \frac{1}{2} a t^2 = \frac{1}{2}(5.20)(32.8)^2 = \boxed{2800 \, \text{m}}
$$

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📌 Notes:

- Always check units — especially converting km/h to m/s.
- Negative acceleration means deceleration (slowing down).
- Use $ v_f = v_i + at $ when you have velocities and time.
- Use $ d = v_i t + \frac{1}{2} a t^2 $ when you need displacement with time and acceleration.

Let me know if you'd like a printed version or help with any other problems!