# Velocity Time graphs: Find acceleration in the first 15 mins in km/h^2

• MHB
In summary, the conversation discusses the concept of uniform acceleration and its application in finding distance and average velocity. The formula $a = \dfrac{v(9.25) - v(9)}{.25 - 0}$ is used to calculate acceleration, and for $9 \le t \le 11$, the distance traveled is determined by the integral $\displaystyle d = \int_9^{11} v(t) \, dt$. The average velocity is represented by the formula $|\bar{v}| = \dfrac{1}{11-9} \int_9^{11} v(t) \, dt$, and is also known as the "area under the curve" in this scenario. The

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(a) note acceleration is uniform from 9:00 to 9:15 ...

$a = \dfrac{v(9.25) - v(9)}{.25 - 0}$

(b) since $v \ge 0$ for $9 \le t \le 11$, $\displaystyle d = \int_9^{11} v(t) \, dt$

(c) $\displaystyle |\bar{v}| = \dfrac{1}{11-9} \int_9^{11} v(t) \, dt$

For (b) and (c) that integral is the "area under the curve". For b, I would think of this as:
1) a triangle with base "1/4 hour" and height "50 km/hr". The area of that triangle is (1/2)(1/4)(50)= 6.25 km.
2) a rectangle with base "1/2 hour" and height "50 km/hr". The area of that rectangle is (1/2)(50)= 25 km.
3) a trapezoid with bases "50 km/hr" and "100 km/hr" and height "1/4 hour" (yes, I've swapped "height" and "base" to better fit the trapezoid). The area is (1/2)(50+ 100)(1/4)= 18.75 km.
4) a rectangle with base "1/2 hr" and height "100 km/hr". The area is (1/2)(100)= 50 km.
5) a triangle with base "1/2 hr" and height "100 km". The area is (1/2)(1/2)(100)= 25 km.

## 1. What is a velocity-time graph?

A velocity-time graph is a visual representation of an object's velocity over time. The horizontal axis represents time, while the vertical axis represents velocity. The slope of the graph at any given point represents the object's acceleration.

## 2. How do you find acceleration on a velocity-time graph?

To find acceleration on a velocity-time graph, you can calculate the slope of the graph at a specific point. This can be done by dividing the change in velocity by the change in time, or by using the formula a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the time interval.

## 3. What is the unit of acceleration on a velocity-time graph?

The unit of acceleration on a velocity-time graph is meters per second squared (m/s^2). This represents the change in velocity over a specific time interval.

## 4. How do you calculate acceleration in the first 15 minutes on a velocity-time graph?

To calculate acceleration in the first 15 minutes on a velocity-time graph, you can first determine the initial and final velocities at the 15-minute mark. Then, you can use the formula a = (vf - vi) / t, where t is the time interval of 15 minutes, to find the acceleration in km/h^2.

## 5. Can acceleration be negative on a velocity-time graph?

Yes, acceleration can be negative on a velocity-time graph. This indicates that the object is slowing down or decelerating. A positive acceleration indicates that the object is speeding up or accelerating.

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