Math / Calculus

QuestionThe figure below shows the velocity $v(t)$ in $\mathrm{ft} / \mathrm{sec}$ of an object moving on the number line, with positive velocities moving to the right.
How far is the object at $t=4$ seconds from its starting point, the origin? The object will be i $\square$ feet to the right of the starting point.

Studdy Solution

STEP 1

1. The velocity-time graph is a triangle, indicating linear changes in velocity.
2. The object starts at the origin, meaning its initial position is 0.
3. The area under the velocity-time graph represents the displacement of the object.

STEP 2

1. Identify the shape and dimensions of the velocity-time graph.
2. Calculate the area under the velocity-time graph to determine displacement.
3. Determine the position of the object at $t = 4$ seconds.

STEP 3

The velocity-time graph is a triangle with a base from $t = 0$ to $t = 4$ seconds and a peak at $t = 2$ seconds with a velocity of 30 ft/sec.

STEP 4

The area of the triangle represents the displacement. The formula for the area of a triangle is:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
In this case, the base is $4$ seconds and the height is $30$ ft/sec.

STEP 5

Substitute the values into the formula:
$\text{Area} = \frac{1}{2} \times 4 \times 30 = 60 \text{ ft}$

STEP 6

The area under the velocity-time graph, which is 60 ft, represents the displacement of the object from its starting point.
The object is $60$ feet to the right of the starting point at $t = 4$ seconds.
The object will be $\boxed{60}$ feet to the right of the starting point.

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Studdy Solution

STEP 1

1. The velocity-time graph is a triangle, indicating linear changes in velocity.
2. The object starts at the origin, meaning its initial position is 0.
3. The area under the velocity-time graph represents the displacement of the object.

STEP 2

1. Identify the shape and dimensions of the velocity-time graph.
2. Calculate the area under the velocity-time graph to determine displacement.
3. Determine the position of the object at $t = 4$ seconds.

STEP 3

The velocity-time graph is a triangle with a base from $t = 0$ to $t = 4$ seconds and a peak at $t = 2$ seconds with a velocity of 30 ft/sec.

STEP 4

The area of the triangle represents the displacement. The formula for the area of a triangle is:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
In this case, the base is $4$ seconds and the height is $30$ ft/sec.

STEP 5

Substitute the values into the formula:
$\text{Area} = \frac{1}{2} \times 4 \times 30 = 60 \text{ ft}$

STEP 6

The area under the velocity-time graph, which is 60 ft, represents the displacement of the object from its starting point.
The object is $60$ feet to the right of the starting point at $t = 4$ seconds.
The object will be $\boxed{60}$ feet to the right of the starting point.

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Studdy Solution

STEP 1

1. The velocity-time graph is a triangle, indicating linear changes in velocity.2. The object starts at the origin, meaning its initial position is 0.3. The area under the velocity-time graph represents the displacement of the object.

STEP 2

1. Identify the shape and dimensions of the velocity-time graph.2. Calculate the area under the velocity-time graph to determine displacement.3. Determine the position of the object at t=4 t = 4 t=4 seconds.

STEP 3

The velocity-time graph is a triangle with a base from t=0 t = 0 t=0 to t=4 t = 4 t=4 seconds and a peak at t=2 t = 2 t=2 seconds with a velocity of 30 ft/sec.

STEP 4

The area of the triangle represents the displacement. The formula for the area of a triangle is:Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}Area=21​×base×heightIn this case, the base is 4 4 4 seconds and the height is 30 30 30 ft/sec.

STEP 5

Substitute the values into the formula:Area=12×4×30=60 ft\text{Area} = \frac{1}{2} \times 4 \times 30 = 60 \text{ ft}Area=21​×4×30=60 ft

STEP 6

The area under the velocity-time graph, which is 60 ft, represents the displacement of the object from its starting point.The object is 60 60 60 feet to the right of the starting point at t=4 t = 4 t=4 seconds. The object will be 60 \boxed{60} 60​ feet to the right of the starting point.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The velocity-time graph is a triangle, indicating linear changes in velocity.
2. The object starts at the origin, meaning its initial position is 0.
3. The area under the velocity-time graph represents the displacement of the object.

1. Identify the shape and dimensions of the velocity-time graph.
2. Calculate the area under the velocity-time graph to determine displacement.
3. Determine the position of the object at $t = 4$ seconds.

The velocity-time graph is a triangle with a base from $t = 0$ to $t = 4$ seconds and a peak at $t = 2$ seconds with a velocity of 30 ft/sec.

The area of the triangle represents the displacement. The formula for the area of a triangle is:
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
In this case, the base is $4$ seconds and the height is $30$ ft/sec.

Substitute the values into the formula:
$\text{Area} = \frac{1}{2} \times 4 \times 30 = 60 \text{ ft}$

The area under the velocity-time graph, which is 60 ft, represents the displacement of the object from its starting point.
The object is $60$ feet to the right of the starting point at $t = 4$ seconds.
The object will be $\boxed{60}$ feet to the right of the starting point.