The Graph above shows the speed of a car traveling in a straight line as a function of time. The car accelerates uniformly and reaches a speed Vb of 4.00 m/s in 8.00 seconds. Calculate the distance traveled by the car from a time of 1.20 to 4.40 seconds.

Well, the car's speedometer seems to be on the move! Let's calculate how much distance it covered during its joyride.

First, let's find the car's initial speed, V0. Since the car accelerates uniformly, we can use the formula:

Vb = V0 + at

where "a" is the acceleration and "t" is the time interval. Rearranging the equation, we have:

V0 = Vb - at

Now, let's key in the values we were given. Vb is 4.00 m/s and t is 8.00 seconds. We need to find "a" first.

Since the car is accelerating uniformly, we can use the formula:

a = (Vb - V0) / t

Substituting the known values, we get:

a = (4.00 m/s - V0) / 8.00 s

Now, let's move on to the fun part – calculating the distance!

The distance traveled during uniform acceleration can be found using the formula:

d = V0*t + (1/2)*a*t^2

We know the time interval, t, is from 1.20 to 4.40 seconds. So, let's plug in the values and get the distance traveled during this time:

d = V0*(4.40 s - 1.20 s) + (1/2)*a*(4.20 s)^2

Now, what's left is to take these formulas to the dance floor of mathematics and let them dance their way to the answer! Remember to substitute the values we found for V0 and a.

And voila! You'll have the distance traveled by our speed-hungry car from 1.20 to 4.40 seconds.

To calculate the distance traveled by the car from a time of 1.20 to 4.40 seconds, we need to find the area under the graph within this time interval.

In the given graph, the speed of the car is constant at 0 m/s until t = 0 seconds. After that, the car accelerates uniformly and reaches a speed of 4.00 m/s in 8.00 seconds.

To find the distance traveled during this time interval, we need to calculate the area of the trapezoid formed by the graph within the time interval of 1.20 to 4.40 seconds.

First, let's find the initial speed of the car at t = 1.20 seconds. Since the car was initially at rest, the initial speed is 0 m/s.

Now, let's use the formula to find the distance traveled during this time interval:

Distance = (V_i + V_f) * t / 2

Where:
- V_i is the initial velocity
- V_f is the final velocity
- t is the time interval

Using the given values:
- V_i = 0 m/s
- V_f = 4.00 m/s
- t = 4.40 s - 1.20 s = 3.20 seconds

Distance = (0 + 4.00) * 3.20 / 2
Distance = 8.00 * 3.20 / 2
Distance = 25.60 / 2
Distance = 12.80 meters

Therefore, the car traveled a distance of 12.80 meters from a time of 1.20 to 4.40 seconds.

To calculate the distance traveled by the car from a time of 1.20 to 4.40 seconds, we can use the area under the velocity-time graph during that time interval. Since the car is accelerating uniformly, the velocity-time graph is a straight line. The distance traveled is equal to the area under this line.

Step 1: Find the equation of the line.
We know that the car reaches a speed of 4.00 m/s in 8.00 seconds. We can use this information to find the equation of the line using the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. Here, u is the initial velocity, which is 0 m/s (since the car starts from rest), and v is 4.00 m/s. We also know that the time taken (t) is 8.00 seconds. Plugging these values into the equation, we get:

4.00 = 0 + a * 8.00

Simplifying the equation:

4.00 = 8.00a

Dividing both sides by 8.00:

a = 0.50 m/s^2

So, the equation of the line is v = 0.50t.

Step 2: Calculate the distance.
Now that we have the equation of the line, we can find the distance traveled by calculating the area under the line. Since the graph represents velocity on the y-axis and time on the x-axis, the area under the line will give us the distance traveled.

To calculate the area, we will integrate the equation of the line with respect to time, over the time interval of 1.20 to 4.40 seconds.

∫[1.20, 4.40] (0.50t) dt

Integrating the function:

[0.50(t^2)/2] [1.20, 4.40]

Simplifying:

[0.25(t^2)] [1.20, 4.40]

Substituting the upper and lower limits:

[0.25(4.40^2) - 0.25(1.20^2)]

Calculating:

[0.25(19.36) - 0.25(1.44)]

[4.84 - 0.36]

4.48 meters

Therefore, the distance traveled by the car from a time of 1.20 to 4.40 seconds is 4.48 meters.

d = 0.5a*t2^2 - 0.5a*t1^2

d = 0.5*0.5*4.4^2 - 0.5*0.5*1.2^2
d = 4.84 - 0.36 = 4.48 m.