Acceleration Worksheet | PDF | Acceleration | Mass - Free Printable

Let's solve each problem step by step using the kinematic equations of motion. The basic equations we’ll use are:

1. $ v = v_0 + at $
2. $ x = x_0 + v_0 t + \frac{1}{2} a t^2 $
3. $ v^2 = v_0^2 + 2a(x - x_0) $

Where:
- $ v $ = final velocity
- $ v_0 $ = initial velocity
- $ a $ = acceleration
- $ t $ = time
- $ x $ = displacement

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Problem 1:

Shelly starts from rest on her bicycle at the top of a hill. After 6.0 s she has reached a final velocity of 14 m/s. What is Shelly’s acceleration?

Given:
- $ v_0 = 0 $ m/s (starts from rest)
- $ v = 14 $ m/s
- $ t = 6.0 $ s

Use equation:
$ v = v_0 + at $

Substitute values:
$$
14 = 0 + a(6.0)
\Rightarrow a = \frac{14}{6.0} = 2.33 \, \text{m/s}^2
$$

✔ Answer: $ \boxed{2.33 \, \text{m/s}^2} $

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Problem 2:

A ball is rolling at 4.80 m/s over level ground when it encounters a ramp, which gives it an acceleration of -0.875 m/s². If the ramp is 0.750 m long, what is the final velocity of the ball when it reaches the top of the ramp?

Given:
- $ v_0 = 4.80 $ m/s
- $ a = -0.875 $ m/s²
- $ x - x_0 = 0.750 $ m
- Find $ v $

Use equation:
$ v^2 = v_0^2 + 2a(x - x_0) $

Substitute:
$$
v^2 = (4.80)^2 + 2(-0.875)(0.750)
= 23.04 - 1.3125 = 21.7275
$$
$$
v = \sqrt{21.7275} \approx 4.66 \, \text{m/s}
$$

✔ Answer: $ \boxed{4.66 \, \text{m/s}} $

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Problem 3:

Bill’s motorcycle can accelerate at 7.05 m/s² at a certain RPM and gear. How far, starting from rest, will Bill travel in the first 2.50 s?

Given:
- $ v_0 = 0 $
- $ a = 7.05 $ m/s²
- $ t = 2.50 $ s
- Find $ x $

Use:
$ x = x_0 + v_0 t + \frac{1}{2} a t^2 $

Since $ x_0 = 0 $, $ v_0 = 0 $:
$$
x = \frac{1}{2}(7.05)(2.50)^2 = \frac{1}{2}(7.05)(6.25) = 22.03 \, \text{m}
$$

✔ Answer: $ \boxed{22.0 \, \text{m}} $ (rounded to 3 significant figures)

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Problem 4:

Lisa drops a ball. If the ball accelerates at 9.80 m/s², how long will it take the ball to reach a velocity of 15.0 m/s?

Given:
- $ v_0 = 0 $ (dropped from rest)
- $ v = 15.0 $ m/s
- $ a = 9.80 $ m/s²
- Find $ t $

Use:
$ v = v_0 + at $

$$
15.0 = 0 + 9.80t \Rightarrow t = \frac{15.0}{9.80} \approx 1.53 \, \text{s}
$$

✔ Answer: $ \boxed{1.53 \, \text{s}} $

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Problem 5:

Big Bob is on his Harley and moving at 14.0 m/s. He then accelerates to a velocity of 25.0 m/s over a distance of 0.250 km. What is Big Bob’s acceleration?

Given:
- $ v_0 = 14.0 $ m/s
- $ v = 25.0 $ m/s
- $ x = 0.250 \, \text{km} = 250 \, \text{m} $
- Find $ a $

Use:
$ v^2 = v_0^2 + 2a(x - x_0) $

$$
(25.0)^2 = (14.0)^2 + 2a(250)
\Rightarrow 625 = 196 + 500a
\Rightarrow 500a = 429
\Rightarrow a = \frac{429}{500} = 0.858 \, \text{m/s}^2
$$

✔ Answer: $ \boxed{0.858 \, \text{m/s}^2} $

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Problem 6:

Chuck’s car is moving at 65.0 m/s when he suddenly accelerates his car at 15.0 m/s² for 3.00 s. How far did Chuck, and his car, travel while he was accelerating?

Given:
- $ v_0 = 65.0 $ m/s
- $ a = 15.0 $ m/s²
- $ t = 3.00 $ s
- Find $ x $

Use:
$ x = x_0 + v_0 t + \frac{1}{2} a t^2 $

Assume $ x_0 = 0 $:
$$
x = 65.0(3.00) + \frac{1}{2}(15.0)(3.00)^2
= 195 + \frac{1}{2}(15.0)(9.00)
= 195 + 67.5 = 262.5 \, \text{m}
$$

✔ Answer: $ \boxed{263 \, \text{m}} $ (rounded to 3 significant figures)

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E.C. Derive the final two formulas for acceleration...

We’re asked to derive the last two kinematic equations from the original three.

The "original three" typically are:
1. $ v = v_0 + at $
2. $ x = x_0 + v_0 t + \frac{1}{2} a t^2 $
3. $ x = x_0 + \frac{1}{2}(v_0 + v)t $

We’ll derive:
- $ v^2 = v_0^2 + 2a(x - x_0) $
- $ x = x_0 + vt - \frac{1}{2}at^2 $ (or equivalent)

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#### Derivation 1: $ v^2 = v_0^2 + 2a(x - x_0) $

Start with:
1. $ v = v_0 + at $ → solve for $ t $:
$ t = \frac{v - v_0}{a} $

2. Use $ x = x_0 + v_0 t + \frac{1}{2} a t^2 $

Substitute $ t $ into this:

$$
x - x_0 = v_0 \left(\frac{v - v_0}{a}\right) + \frac{1}{2} a \left(\frac{v - v_0}{a}\right)^2
$$

$$
= \frac{v_0(v - v_0)}{a} + \frac{1}{2} a \cdot \frac{(v - v_0)^2}{a^2}
= \frac{v_0(v - v_0)}{a} + \frac{(v - v_0)^2}{2a}
$$

Multiply both sides by $ 2a $ to eliminate denominators:

$$
2a(x - x_0) = 2v_0(v - v_0) + (v - v_0)^2
$$

Expand:
$$
= 2v_0 v - 2v_0^2 + v^2 - 2v v_0 + v_0^2
= v^2 - v_0^2
$$

So:
$$
2a(x - x_0) = v^2 - v_0^2 \Rightarrow v^2 = v_0^2 + 2a(x - x_0)
$$

✔ Derived!

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#### Derivation 2: Alternative form — $ x = x_0 + vt - \frac{1}{2} a t^2 $

We know:
$ v = v_0 + at \Rightarrow v_0 = v - at $

Now plug into:
$ x = x_0 + v_0 t + \frac{1}{2} a t^2 $

$$
x = x_0 + (v - at)t + \frac{1}{2} a t^2
= x_0 + vt - a t^2 + \frac{1}{2} a t^2
= x_0 + vt - \frac{1}{2} a t^2
$$

✔ So: $ x = x_0 + vt - \frac{1}{2} a t^2 $

This is the second derived formula.

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✔ Final Answers Summary:

1. $ \boxed{2.33 \, \text{m/s}^2} $
2. $ \boxed{4.66 \, \text{m/s}} $
3. $ \boxed{22.0 \, \text{m}} $
4. $ \boxed{1.53 \, \text{s}} $
5. $ \boxed{0.858 \, \text{m/s}^2} $
6. $ \boxed{263 \, \text{m}} $

And the two derived formulas:
- $ v^2 = v_0^2 + 2a(x - x_0) $
- $ x = x_0 + vt - \frac{1}{2} a t^2 $

These are now ready to be written on the back of the worksheet!