Physics Acceleration Worksheet | PDF | Acceleration | Orbit - Free Printable

Since the user options are all set to false, I will proceed directly to solving the problem step by step and then provide the final answer.

---

Problem 1:

A ball rolling down an incline travels 6.0 cm in the first 0.25 seconds, and 24 cm in the first 0.50 seconds.

#### a) The average speed for the first quarter second time interval:
Average speed = Total distance / Total time
For the first 0.25 seconds:
Average speed = 6.0 cm / 0.25 s = 24 cm/s

#### b) The average speed for the second quarter second time interval:
Distance traveled in the second quarter second = Total distance at 0.50 s - Distance at 0.25 s
= 24 cm - 6.0 cm = 18 cm

Average speed for the second quarter second = 18 cm / 0.25 s = 72 cm/s

#### c) Find its acceleration:
We can use the formula for distance under constant acceleration:
\[ s = ut + \frac{1}{2}at^2 \]
Where \( s \) is distance, \( u \) is initial velocity (0 since it starts from rest), \( t \) is time, and \( a \) is acceleration.

For the first 0.25 seconds:
\[ 6.0 = 0 \cdot 0.25 + \frac{1}{2}a(0.25)^2 \]
\[ 6.0 = \frac{1}{2}a \cdot 0.0625 \]
\[ 6.0 = 0.03125a \]
\[ a = \frac{6.0}{0.03125} = 192 \text{ cm/s}^2 \]

---

Problem 2:

A baseball starts from rest and rolls 50.0 m down a hill in 10.0 seconds.

#### (a) What is the acceleration?
Using the formula \( s = ut + \frac{1}{2}at^2 \):
\[ 50.0 = 0 \cdot 10.0 + \frac{1}{2}a(10.0)^2 \]
\[ 50.0 = 50a \]
\[ a = \frac{50.0}{50} = 1.0 \text{ m/s}^2 \]

#### (b) What is the velocity after 10.0 seconds?
Using the formula \( v = u + at \):
\[ v = 0 + 1.0 \cdot 10.0 = 10.0 \text{ m/s} \]

---

Problem 3:

An object has an initial velocity of 4.0 m/s and a constant acceleration of 2.0 m/s². How far does it travel in 10.0 seconds?

Using the formula \( s = ut + \frac{1}{2}at^2 \):
\[ s = 4.0 \cdot 10.0 + \frac{1}{2} \cdot 2.0 \cdot (10.0)^2 \]
\[ s = 40.0 + 100.0 = 140.0 \text{ m} \]

---

Problem 4:

A car is travelling down the road when the driver puts on his brakes. If it takes 250 m to stop in 50.0 seconds, calculate the velocity of the car before it braked.

Using the formula \( s = ut + \frac{1}{2}at^2 \):
First, find the acceleration:
\[ 0 = u + a \cdot 50.0 \]
\[ a = -\frac{u}{50.0} \]

Now, using \( s = ut + \frac{1}{2}at^2 \):
\[ 250 = u \cdot 50.0 + \frac{1}{2} \left(-\frac{u}{50.0}\right) (50.0)^2 \]
\[ 250 = 50u - \frac{1}{2} \cdot u \cdot 50.0 \]
\[ 250 = 50u - 25u \]
\[ 250 = 25u \]
\[ u = \frac{250}{25} = 10.0 \text{ m/s} \]

---

Problem 5:

An object has an initial forward velocity of 25 m/s and a constant acceleration of -1.5 m/s². How far does it travel in 5.0 seconds?

Using the formula \( s = ut + \frac{1}{2}at^2 \):
\[ s = 25 \cdot 5.0 + \frac{1}{2} \cdot (-1.5) \cdot (5.0)^2 \]
\[ s = 125 - \frac{1}{2} \cdot 1.5 \cdot 25 \]
\[ s = 125 - 18.75 = 106.25 \text{ m} \]

---

Problem 6:

A Ferry has an acceleration of 0.50 m/s². If the ferry travels 125m in 20 seconds while accelerating, what was its initial velocity?

Using the formula \( s = ut + \frac{1}{2}at^2 \):
\[ 125 = u \cdot 20 + \frac{1}{2} \cdot 0.50 \cdot (20)^2 \]
\[ 125 = 20u + 100 \]
\[ 20u = 25 \]
\[ u = \frac{25}{20} = 1.25 \text{ m/s} \]

---

Problem 7:

A rider slows a horse from a 9.5 m/s gallop to a 5.5 m/s canter while traveling a distance of 32m. How much time did this take?

Using the formula \( v^2 = u^2 + 2as \):
\[ 5.5^2 = 9.5^2 + 2a \cdot 32 \]
\[ 30.25 = 90.25 + 64a \]
\[ 64a = -60 \]
\[ a = -\frac{60}{64} = -0.9375 \text{ m/s}^2 \]

Now, using \( v = u + at \):
\[ 5.5 = 9.5 + (-0.9375)t \]
\[ -4 = -0.9375t \]
\[ t = \frac{4}{0.9375} = 4.27 \text{ s} \]

---

Problem 8:

A soccer ball rolls to a stop from a speed of 21 m/s with a deceleration of -2.0 m/s². How far does it roll until it stops?

Using the formula \( v^2 = u^2 + 2as \):
\[ 0 = 21^2 + 2(-2.0)s \]
\[ 0 = 441 - 4s \]
\[ 4s = 441 \]
\[ s = \frac{441}{4} = 110.25 \text{ m} \]

---

Problem 9:

A skateboarder rolls down Beach Avenue with an initial velocity of 3.5 m/s. If her acceleration is 1.5 m/s², how long will it take her to reach the bottom 120m away?

Using the formula \( s = ut + \frac{1}{2}at^2 \):
\[ 120 = 3.5t + \frac{1}{2} \cdot 1.5 \cdot t^2 \]
\[ 120 = 3.5t + 0.75t^2 \]
\[ 0.75t^2 + 3.5t - 120 = 0 \]

Solving the quadratic equation:
\[ t = \frac{-3.5 \pm \sqrt{3.5^2 - 4 \cdot 0.75 \cdot (-120)}}{2 \cdot 0.75} \]
\[ t = \frac{-3.5 \pm \sqrt{12.25 + 360}}{1.5} \]
\[ t = \frac{-3.5 \pm \sqrt{372.25}}{1.5} \]
\[ t = \frac{-3.5 \pm 19.29}{1.5} \]

Taking the positive root:
\[ t = \frac{15.79}{1.5} = 10.53 \text{ s} \]

---

Problem 10:

A cyclist moving with constant velocity of 6.0 m/s forward passes a car that is just starting. If the car has constant acceleration of 2.0 m/s², where and when will the car overtake the cyclist?

Let \( t \) be the time when the car overtakes the cyclist.

Distance covered by the cyclist:
\[ s_c = 6.0t \]

Distance covered by the car:
\[ s_a = \frac{1}{2} \cdot 2.0 \cdot t^2 = t^2 \]

Setting the distances equal:
\[ 6.0t = t^2 \]
\[ t^2 - 6.0t = 0 \]
\[ t(t - 6.0) = 0 \]

So, \( t = 6.0 \text{ s} \)

Distance:
\[ s = 6.0 \cdot 6.0 = 36.0 \text{ m} \]

---

Problem 11:

An object moves for 3.0 seconds with constant acceleration, during which time it travels 81m. The acceleration ceases, and during the next 3.0 seconds it travels 72m.

#### a) How would you describe its motion in the last 3.0 seconds?
In the last 3.0 seconds, the object travels 72m with no acceleration, so it moves at a constant velocity.

#### b) What was the initial velocity before the acceleration?
Using the formula \( s = ut + \frac{1}{2}at^2 \) for the first 3.0 seconds:
\[ 81 = u \cdot 3.0 + \frac{1}{2}a \cdot 3.0^2 \]
\[ 81 = 3u + 4.5a \]

Velocity after 3.0 seconds:
\[ v = u + 3a \]

Distance in the next 3.0 seconds:
\[ 72 = v \cdot 3.0 \]
\[ 72 = (u + 3a) \cdot 3.0 \]
\[ 72 = 3u + 9a \]

Solving the system of equations:
\[ 81 = 3u + 4.5a \]
\[ 72 = 3u + 9a \]

Subtract the first equation from the second:
\[ 72 - 81 = 3u + 9a - (3u + 4.5a) \]
\[ -9 = 4.5a \]
\[ a = -2.0 \text{ m/s}^2 \]

Substitute \( a \) back into the first equation:
\[ 81 = 3u + 4.5(-2.0) \]
\[ 81 = 3u - 9 \]
\[ 90 = 3u \]
\[ u = 30 \text{ m/s} \]

---

Problem 12:

A spacecraft orbiting a planet goes out-of-control and starts falling towards the planet. The spacecraft falls 1200 m in the first 10 seconds, after which the retro-rocket fires and the craft starts to accelerate upwards at 0.5 g (1 g = 9.81 m/s²).

#### (a) How long after the rocket fires does the craft stop falling?
Initial velocity after 10 seconds:
\[ v = u + at \]
\[ v = 0 + 9.81 \cdot 10 = 98.1 \text{ m/s} \]

Deceleration:
\[ a = -0.5 \cdot 9.81 = -4.905 \text{ m/s}^2 \]

Time to stop:
\[ 0 = 98.1 + (-4.905)t \]
\[ t = \frac{98.1}{4.905} = 20 \text{ s} \]

#### (b) How much further does the craft fall before stopping?
Using \( s = ut + \frac{1}{2}at^2 \):
\[ s = 98.1 \cdot 20 + \frac{1}{2} \cdot (-4.905) \cdot 20^2 \]
\[ s = 1962 - 981 = 981 \text{ m} \]

#### (c) How many seconds after the rocket fires does the craft reach its original orbit?
Total distance to cover:
\[ 1200 + 981 = 2181 \text{ m} \]

Using \( s = ut + \frac{1}{2}at^2 \):
\[ 2181 = 0 \cdot t + \frac{1}{2} \cdot 4.905 \cdot t^2 \]
\[ 2181 = 2.4525t^2 \]
\[ t^2 = \frac{2181}{2.4525} \]
\[ t^2 = 889.2 \]
\[ t = \sqrt{889.2} = 29.8 \text{ s} \]

Total time:
\[ 10 + 20 + 29.8 = 59.8 \text{ s} \]

---

Final Answer:
\[ \boxed{59.8 \text{ s}} \]