Suppose a hovercraft undergoes constant uniform acceleration, which causes the hovercraft to move from rest to a speed of 30.8 m/s. How long does the hovercraft accelerate if it travels a distance of 493 m?
To find the time it takes for the hovercraft to accelerate and travel a certain distance, we can use the kinematic equation:
\[ d = v_0t + \frac{1}{2}at^2 \]
where:
- \( d \) is the distance traveled (493 m)
- \( v_0 \) is the initial velocity (0 m/s since it starts from rest)
- \( a \) is the constant acceleration
- \( t \) is the time it takes to accelerate and travel the distance
We need to rearrange the equation to solve for \( t \). First, let's simplify the equation by substituting the known values:
\[ 493 = 0 \times t + \frac{1}{2}a t^2 \]
Since \( 0 \times t \) is 0, the equation simplifies to:
\[ 493 = \frac{1}{2}a t^2 \]
Now, let's isolate \( t \). We can do this by dividing both sides of the equation by \( \frac{1}{2}a \):
\[ \frac{493}{\frac{1}{2}a} = t^2 \]
To simplify further, we can multiply the numerator and denominator by 2:
\[ 2 \times 493 = t^2 \times a \]
\[ 986 = t^2 \times a \]
Finally, we can take the square root of both sides to solve for \( t \):
\[ t = \sqrt{\frac{986}{a}} \]
To find the value of \( a \), we use the equation for constant acceleration:
\[ v = v_0 + at \]
We're given that the final velocity, \( v \), is 30.8 m/s. Since the initial velocity, \( v_0 \), is 0 m/s, the equation simplifies to:
\[ v = at \]
Rearranging the equation to solve for \( a \):
\[ a = \frac{v}{t} \]
Now we can substitute this value of \( a \) into the equation for \( t \):
\[ t = \sqrt{\frac{986}{\frac{v}{t}}} \]
Simplifying the equation:
\[ t = \sqrt{\frac{986t}{v}} \]
To isolate \( t \), let's square both sides:
\[ t^2 = \frac{986t}{v} \]
Rearranging the equation:
\[ t^2v = 986t \]
Divide both sides by \( t \) (assuming \( t \neq 0 \)):
\[ tv = 986 \]
Solving for \( t \):
\[ t = \frac{986}{v} \]
Now we have the expression for \( t \) in terms of \( v \). Substitute the given value of \( v \) (30.8 m/s) into this equation to find the time it takes for the hovercraft to accelerate and travel the distance of 493 m:
\[ t = \frac{986}{30.8} \]
\[ t \approx 32.03 \, \text{s} \]
Therefore, the hovercraft accelerates for approximately 32.03 seconds to travel a distance of 493 m.