Solved Equations of Motion Worksheet 1. A car moving at a | Chegg.com - Free Printable

Equations of Motion Worksheet Solutions

The equations of motion we will use are:
1. \( v = u + at \)
2. \( s = ut + \frac{1}{2}at^2 \)
3. \( v^2 = u^2 + 2as \)

Where:
- \( u \) = initial velocity
- \( v \) = final velocity
- \( a \) = acceleration
- \( t \) = time
- \( s \) = displacement

Let's solve each problem step by step.

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Problem 1

A car moving at a velocity of 25 m/s accelerates at a rate of 6 m/s². Find its velocity after 3 s.

Given:
- Initial velocity, \( u = 25 \) m/s
- Acceleration, \( a = 6 \) m/s²
- Time, \( t = 3 \) s

Formula to use:
\[ v = u + at \]

Substitute the values:
\[ v = 25 + (6)(3) \]
\[ v = 25 + 18 \]
\[ v = 43 \] m/s

Answer:
\[ \boxed{43} \]

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Problem 2

An object is dropped from rest. Calculate its velocity after 2.5 s if it is dropped:
a. On Earth, where the acceleration due to gravity is 9.8 m/s².
b. On Mars, where the acceleration due to gravity is 3.8 m/s².

Given:
- Initial velocity, \( u = 0 \) m/s (dropped from rest)
- Time, \( t = 2.5 \) s

#### Part (a): On Earth
- Acceleration, \( a = 9.8 \) m/s²

Formula to use:
\[ v = u + at \]

Substitute the values:
\[ v = 0 + (9.8)(2.5) \]
\[ v = 24.5 \] m/s

Answer for part (a):
\[ \boxed{24.5} \]

#### Part (b): On Mars
- Acceleration, \( a = 3.8 \) m/s²

Formula to use:
\[ v = u + at \]

Substitute the values:
\[ v = 0 + (3.8)(2.5) \]
\[ v = 9.5 \] m/s

Answer for part (b):
\[ \boxed{9.5} \]

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Problem 3

A motorbike is traveling with a velocity of 3 m/s. It accelerates at a rate of 9.3 m/s² for 1.8 s. Calculate the distance it travels in this time.

Given:
- Initial velocity, \( u = 3 \) m/s
- Acceleration, \( a = 9.3 \) m/s²
- Time, \( t = 1.8 \) s

Formula to use:
\[ s = ut + \frac{1}{2}at^2 \]

Substitute the values:
\[ s = (3)(1.8) + \frac{1}{2}(9.3)(1.8)^2 \]
\[ s = 5.4 + \frac{1}{2}(9.3)(3.24) \]
\[ s = 5.4 + \frac{1}{2}(30.012) \]
\[ s = 5.4 + 15.006 \]
\[ s = 20.406 \] m

Answer:
\[ \boxed{20.4} \]

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Problem 4

A Tesla Roadster car accelerates from rest at a rate of 7.1 m/s² for a time of 3.9 s. Calculate the distance it travels in this time.

Given:
- Initial velocity, \( u = 0 \) m/s (from rest)
- Acceleration, \( a = 7.1 \) m/s²
- Time, \( t = 3.9 \) s

Formula to use:
\[ s = ut + \frac{1}{2}at^2 \]

Substitute the values:
\[ s = (0)(3.9) + \frac{1}{2}(7.1)(3.9)^2 \]
\[ s = 0 + \frac{1}{2}(7.1)(15.21) \]
\[ s = \frac{1}{2}(108.091) \]
\[ s = 54.0455 \] m

Answer:
\[ \boxed{54.0} \]

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Problem 5

A bullet accelerates at a rate of 90,000 m/s² from rest. Calculate its velocity if it travels a distance of 0.5 m while accelerating.

Given:
- Initial velocity, \( u = 0 \) m/s (from rest)
- Acceleration, \( a = 90,000 \) m/s²
- Distance, \( s = 0.5 \) m

Formula to use:
\[ v^2 = u^2 + 2as \]

Substitute the values:
\[ v^2 = (0)^2 + 2(90,000)(0.5) \]
\[ v^2 = 0 + 90,000 \]
\[ v^2 = 90,000 \]
\[ v = \sqrt{90,000} \]
\[ v = 300 \] m/s

Answer:
\[ \boxed{300} \]

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Problem 6

An aircraft is traveling along a runway at a velocity of 25 m/s. It accelerates at a rate of 4 m/s² for a distance of 750 m before taking off. Calculate its take-off speed.

Given:
- Initial velocity, \( u = 25 \) m/s
- Acceleration, \( a = 4 \) m/s²
- Distance, \( s = 750 \) m

Formula to use:
\[ v^2 = u^2 + 2as \]

Substitute the values:
\[ v^2 = (25)^2 + 2(4)(750) \]
\[ v^2 = 625 + 6000 \]
\[ v^2 = 6625 \]
\[ v = \sqrt{6625} \]
\[ v \approx 81.4 \] m/s

Answer:
\[ \boxed{81.4} \]

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Problem 7

A car is traveling at a speed of 21 m/s. It accelerates at an average rate of 3 m/s² for a time of 4 seconds. Find the distance it travels.

Given:
- Initial velocity, \( u = 21 \) m/s
- Acceleration, \( a = 3 \) m/s²
- Time, \( t = 4 \) s

Formula to use:
\[ s = ut + \frac{1}{2}at^2 \]

Substitute the values:
\[ s = (21)(4) + \frac{1}{2}(3)(4)^2 \]
\[ s = 84 + \frac{1}{2}(3)(16) \]
\[ s = 84 + \frac{1}{2}(48) \]
\[ s = 84 + 24 \]
\[ s = 108 \] m

Answer:
\[ \boxed{108} \]

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Problem 8

A car accelerates at a rate of 10 m/s² for a time of 4 s. It reaches a speed of 52 m/s. Calculate its initial speed.

Given:
- Final velocity, \( v = 52 \) m/s
- Acceleration, \( a = 10 \) m/s²
- Time, \( t = 4 \) s

Formula to use:
\[ v = u + at \]

Rearrange to solve for \( u \):
\[ u = v - at \]

Substitute the values:
\[ u = 52 - (10)(4) \]
\[ u = 52 - 40 \]
\[ u = 12 \] m/s

Answer:
\[ \boxed{12} \]

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Problem 9

A bullet is at rest. It travels a distance of 0.34 m in a time of 0.0095 seconds. Calculate its acceleration.

Given:
- Initial velocity, \( u = 0 \) m/s (at rest)
- Distance, \( s = 0.34 \) m
- Time, \( t = 0.0095 \) s

Formula to use:
\[ s = ut + \frac{1}{2}at^2 \]

Since \( u = 0 \):
\[ s = \frac{1}{2}at^2 \]

Rearrange to solve for \( a \):
\[ a = \frac{2s}{t^2} \]

Substitute the values:
\[ a = \frac{2(0.34)}{(0.0095)^2} \]
\[ a = \frac{0.68}{0.00009025} \]
\[ a \approx 7536.5 \] m/s²

Answer:
\[ \boxed{7537} \]

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Problem 10

A falcon is diving at a speed of 30 m/s. It accelerates at a rate of 40 m/s² and reaches a speed of 100 m/s. Calculate the distance it travels during its acceleration.

Given:
- Initial velocity, \( u = 30 \) m/s
- Final velocity, \( v = 100 \) m/s
- Acceleration, \( a = 40 \) m/s²

Formula to use:
\[ v^2 = u^2 + 2as \]

Rearrange to solve for \( s \):
\[ s = \frac{v^2 - u^2}{2a} \]

Substitute the values:
\[ s = \frac{(100)^2 - (30)^2}{2(40)} \]
\[ s = \frac{10000 - 900}{80} \]
\[ s = \frac{9100}{80} \]
\[ s = 113.75 \] m

Answer:
\[ \boxed{113.8} \]

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Final Answers:

1. \( \boxed{43} \)
2. (a) \( \boxed{24.5} \), (b) \( \boxed{9.5} \)
3. \( \boxed{20.4} \)
4. \( \boxed{54.0} \)
5. \( \boxed{300} \)
6. \( \boxed{81.4} \)
7. \( \boxed{108} \)
8. \( \boxed{12} \)
9. \( \boxed{7537} \)
10. \( \boxed{113.8} \)