During takeoff, an airplane goes from 0 to 78 m/s in 13 s what is the acceleration?
How fast is it going after 6 s?
How far has it traveled by the time it reaches 78 m/s
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a = ∆v/∆t = ((78-0)m/s)/(13s) = 6 m/s^2
v = at
s = 1/2 at^2
...
To calculate the acceleration, you can use the formula:
\[ acceleration = \frac{{final\ velocity - initial\ velocity}}{{time}} \]
In this case, the initial velocity is 0 m/s, the final velocity is 78 m/s, and the time is 13 s. Plugging these values into the formula, we get:
\[ acceleration = \frac{{78\ m/s - 0\ m/s}}{{13\ s}} \]
So, the acceleration is:
\[ acceleration = \frac{{78}}{{13}} \]
To find how fast the airplane is going after 6 seconds, you can use the acceleration formula along with the initial velocity of 0 m/s and the time of 6 s:
\[ final\ velocity = initial\ velocity + (acceleration \times time) \]
Plugging in the values, we get:
\[ final\ velocity = 0\ m/s + \left(\frac{{78}}{{13}} \times 6\ s\right) \]
To calculate it:
\[ final\ velocity = \left(\frac{{78}}{{13}} \times 6\right) \]
To find the distance traveled by the time the airplane reaches 78 m/s, you can use the formula:
\[ distance = \frac{1}{2} \times acceleration \times time^2 + initial\ velocity \times time \]
In this case, the acceleration is the same as before, the time is unknown, and the initial velocity is 0 m/s. We want to solve for time, so we rearrange the formula:
\[ \frac{1}{2} \times acceleration \times time^2 + initial\ velocity \times time - distance = 0 \]
Using the quadratic formula, we can solve for time:
\[ time = \frac{{-b \pm \sqrt{{b^2 -4ac}}}}{2a} \]
Where:
\[ a = \frac{1}{2} \times acceleration \]
\[ b = initial\ velocity \]
\[ c = -distance \]
Plugging in the values, we get:
\[ time = \frac{{-initial\ velocity \pm \sqrt{{initial\ velocity^2 -4 \times (1/2) \times acceleration \times -distance}}}}{2 \times (1/2) \times acceleration} \]
To calculate it, you can plug in the values for initial velocity, acceleration, and distance into this formula and solve for time.