Question

4. A jet plane accelerates uniformly starting from rest to attain a speed 216 km/hour in 1minute. Calculate the (i) acceleration and (ii) the distance covered by the jet plane in that time.​

Answer 1

From question, we have;

• Intial velocity of jet = 0 m/s
• Final velocity of jet = 216 km/h = 60 m/s
• Total time taken = 1 min = 60 sec

Now, from second equation of motion;

• v = u + at
• 60 = 0 + a(60)
• 60 = 60a
• 60/60 = a
• 1 m/s² = a

Therefore, the acceleration of the jet is 1 m/s².

Atlast, using the third equation of motion;

2as = v² - u²

2(1)s = (60)² - (0)²

2s = 3600

s = 3600/2

s = 1800 m

Therefore, the distance covered by the jet is 1800 m.

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