A ball is rolled horizontally off the top of a table. After 3 seconds the ball lands on the ground with a final velocity of −3.5 m/s. Which kinematic equation would be most useful for finding the ball’s initial velocity? (Assume a= −9.8 m/s2)

Answer 1

To find the ball's initial velocity after it is rolled off the table horizontally, we want to use a kinematic equation that relates the initial velocity (\(v_0\)), final velocity (\(v\)), acceleration (\(a\)), and time (\(t\)).

The best kinematic equation for this situation is:

\[ v = v_0 + at \]

Where:

• \(v\) is the final velocity,
• \(v_0\) is the initial velocity,
• \(a\) is the acceleration, and
• \(t\) is the time duration.

In this case:

• The final vertical velocity (\(v\)) after 3 seconds is \(-3.5 , \text{m/s}\) (the negative indicates downward velocity).
• The acceleration due to gravity (\(a\)) is \(-9.8 , \text{m/s}^2\).
• The time (\(t\)) is \(3 , \text{s}\).

You can rearrange the equation to solve for \(v_0\):

\[ v_0 = v - at \]

Substituting in the values gives:

\[ v_0 = -3.5 , \text{m/s} - (-9.8 , \text{m/s}^2 \cdot 3 , \text{s}) \]

Calculating this will yield the initial horizontal velocity of the ball.

Answer 2

just the answer

Answer 3

The kinematic equation to use is \( v = v_0 + at \).

Answer 4

solve this

Answer 5

Sure! Let's solve using the equation:

\[ v = v_0 + at \]

Given:

• \( v = -3.5 , \text{m/s} \)
• \( a = -9.8 , \text{m/s}^2 \)
• \( t = 3 , \text{s} \)

Rearranging the equation to solve for \( v_0 \):

\[ v_0 = v - at \]

Now substituting in the known values:

\[ v_0 = -3.5 , \text{m/s} - (-9.8 , \text{m/s}^2 \cdot 3 , \text{s}) \]

Calculating \( -9.8 , \text{m/s}^2 \cdot 3 , \text{s} = -29.4 , \text{m/s} \):

\[ v_0 = -3.5 , \text{m/s} + 29.4 , \text{m/s} \]

\[ v_0 = 25.9 , \text{m/s} \]

So, the initial horizontal velocity \( v_0 \) is 25.9 m/s.

Answer 6

solve this

A toy race car zooms across the ground with an acceleration of 2.8 m/s^2. After 4 seconds it has a final velocity of 22.2 m/s, what is the toy car's initial velocity?

Answer 7

To find the toy car's initial velocity, we can use the kinematic equation:

\[ v = v_0 + at \]

Where:

• \( v \) is the final velocity (22.2 m/s),
• \( v_0 \) is the initial velocity,
• \( a \) is the acceleration (2.8 m/s²),
• \( t \) is the time (4 s).

Rearranging the equation to solve for \( v_0 \):

\[ v_0 = v - at \]

Now, substituting in the known values:

\[ v_0 = 22.2 , \text{m/s} - (2.8 , \text{m/s}^2 \cdot 4 , \text{s}) \]

Calculating \( 2.8 , \text{m/s}^2 \cdot 4 , \text{s} \):

\[ 2.8 , \text{m/s}^2 \cdot 4 , \text{s} = 11.2 , \text{m/s} \]

Now substituting this back into the equation:

\[ v_0 = 22.2 , \text{m/s} - 11.2 , \text{m/s} \]

\[ v_0 = 11.0 , \text{m/s} \]

So, the toy car's initial velocity is 11.0 m/s.

Answer 8

If you drive your car at a CONSTANT speed of 30mph for 35 seconds, what is the acceleration for the trip?

Answer 9

When a car is traveling at a constant speed, its acceleration is zero.

Acceleration is defined as the change in velocity over time. Since the speed is constant (30 mph) and there is no change in velocity, the acceleration can be calculated as follows:

\[ \text{Acceleration} (a) = \frac{\Delta v}{\Delta t} \]

where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time.

In this case:

• \(\Delta v = 0\) (since the speed remains constant)
• \(\Delta t = 35\) seconds

Thus:

\[ a = \frac{0}{35 , \text{s}} = 0 , \text{mph}^2 \]

So, the acceleration for the trip is 0 mph².

Answer 10

An Amazon delivery box slides across the floor of the delivery truck and accelerates from 6 m/s to 20 m/s in 1.3 seconds. What is the acceleration of the box?

Answer 11

To find the acceleration of the box, we can use the formula for acceleration:

\[ a = \frac{\Delta v}{\Delta t} \]

Where:

• \( \Delta v \) is the change in velocity,
• \( \Delta t \) is the change in time.

In this case:

• Initial velocity (\( v_0 \)) = 6 m/s
• Final velocity (\( v \)) = 20 m/s
• Time (\( \Delta t \)) = 1.3 seconds

First, calculate the change in velocity (\( \Delta v \)):

\[ \Delta v = v - v_0 = 20 , \text{m/s} - 6 , \text{m/s} = 14 , \text{m/s} \]

Now, substitute the values into the acceleration formula:

\[ a = \frac{14 , \text{m/s}}{1.3 , \text{s}} \approx 10.77 , \text{m/s}^2 \]

Thus, the acceleration of the box is approximately 10.77 m/s².