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Questions and Answers
What is the take off speed of the jet in km/hr?
278 km/hr
What is the distance over which the catapult accelerates the jet in metres?
94 metres
What is the formula that relates speed, acceleration, and time?
v = at
What is the formula for distance covered with constant acceleration?
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If the jet's take off speed is 278 km/hr, what is the speed in metres per second?
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How do you calculate the time taken for the fighter to be accelerated to take off speed?
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Study Notes
Aircraft Carrier Catapult Acceleration Calculation
- Acceleration can be calculated using the final speed and distance available for the jet to reach takeoff speed.
- Given take-off speed: 278 km/hr; convert to m/s for calculations (1 km/hr = 1/3.6 m/s).
- Conversion: 278 km/hr = 278 / 3.6 ≈ 77.22 m/s.
Distance and acceleration
- Distance over which the catapult accelerates the jet: 94 m.
- Use the formula ( S = \frac{1}{2} a t^2 ) rearranged gives acceleration ( a = \frac{2S}{t^2} ).
- To find time ( t ), use ( v = at ) which can be rewritten as ( t = \frac{v}{a} ).
Acceleration Calculation
- When acceleration is constant: combine both equations to eliminate the variable ( t ).
- Substitute ( t ) from the second equation into the first, leading to ( S = \frac{v^2}{2a} ), allowing the calculation of acceleration ( a = \frac{v^2}{2S} ).
- Calculating:
- ( a = \frac{(77.22)^2}{2 \times 94} ).
- Resulting in ( a ≈ 36.87 , \text{m/s}^2 ).
Time Calculation
- After finding acceleration, calculate time:
- Substitute back to find ( t ) using ( t = \frac{v}{a} ).
- ( t = \frac{77.22}{36.87} ).
- Resulting in ( t ≈ 2.09 , \text{seconds} ).
Summary
- Final acceleration of the jet is approximately 36.87 m/s².
- Time taken for the jet to reach take-off speed is approximately 2.09 seconds.
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Description
This quiz covers the calculations involved in determining the acceleration of a jet using an aircraft carrier catapult. You'll learn to convert speeds, use kinematic equations, and calculate both acceleration and time. Test your knowledge on the physics behind jet take-off dynamics.
