Q.47 A jet plane lands on an aircraft carrier at 70 m/s and stops in 3 seconds. Assuming that the acceleration is constant, the jet plane travels a distance of _____ m before it stops.
Problem Solution
The jet has initial velocity \(u = 70\) m/s, final velocity \(v = 0\) m/s, and time \(t = 3\) s.
First, calculate acceleration using \(v = u + at\):
\[ a = \frac{v – u}{t} = \frac{0 – 70}{3} = -23.33 \, \text{m/s}^2 \]
Then, find distance using \(s = ut + \frac{1}{2}at^2\):
\[ s = (70)(3) + \frac{1}{2}(-23.33)(3)^2 = 210 – 105 = 105 \, \text{m} \]
Alternatively, average velocity \(\frac{u + v}{2} = 35\) m/s gives \(s = 35 \times 3 = 105\) m.
Key Physics Formulas
- Acceleration: \(a = \frac{v – u}{t}\) relates change in velocity to time.
- Distance: \(s = \frac{u + v}{2} t\) works for constant acceleration (arithmetic mean velocity).
- Third equation: \(v^2 = u^2 + 2as\) yields \(s = \frac{u^2}{2|a|} = \frac{70^2}{2 \times 23.33} = 105\) m, confirming consistency.
SEO Optimized Article
Jet plane aircraft carrier landing at 70 m/s presents a classic kinematics problem for physics students preparing for CSIR NET or IIT JAM exams. This scenario tests constant acceleration principles during high-speed carrier deck arrests, where jets decelerate rapidly using arrestor wires.
Step-by-Step Calculation
- Identify givens: Initial speed \(u = 70\) m/s, final speed \(v = 0\), time \(t = 3\) s.
- Compute deceleration: \(a = -23.33\) m/s² (about 2.4g, realistic for hydraulic arrestors).
- Distance traveled: 105 m, matching real carrier deck limits (around 100-300 m).
Real-world carriers use arrestor wires for such stops, achieving 20-30 m/s² deceleration over short distances.
Common Exam Mistakes
- Forgetting negative acceleration sign (deceleration).
- Using \(s = ut\) alone (ignores deceleration effect) — gives 210 m (wrong).
- No options provided, but typical MCQs might list 70, 105, 140, 210 m—210 m is initial \(ut\), incorrect without \(\frac{1}{2}at^2\) correction.
