A 747 commercial jet liner moves down a runway at a constant acceleration,starting from rest and reaching the take-off speed of 285 km/hr after travelling 3400 m down the runway.The acceleration of the jet on the runway is

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Quizplus · Accepted Answer

To find the acceleration, use the formula $ v^2 = u^2 + 2as $. Here, $ v = 285 \times \frac{1000}{3600} \, \text{m/s} $, $ u = 0 $, and $ s = 3400 \, \text{m} $. Solving for $ a $ gives approximately $ 0.922 \, \text{m/s}^2 $.

A 747 Commercial Jet Liner Moves Down a Runway at a Constant