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.Correct answer is between '105,105'. Can you explain this answer?

Question

Answer

Solution:

To find the distance traveled by the jet plane before it stops, we need to use the formula:

distance = (initial velocity * time) + 1/2 (acceleration * time^2)

Given data:

Initial velocity (u) = 70 m/s
Time (t) = 3 s
Acceleration (a) = ?

We know that the jet plane comes to a stop, so the final velocity (v) = 0 m/s.

Using the equation of motion:

v = u + at

0 = 70 + a * 3

a = -70/3 m/s^2 (negative sign indicates deceleration)

Substituting the values of u, t, and a in the formula for distance:

distance = (70 * 3) + 1/2 (-70/3) * (3^2)

distance = 210 - 70 = 140 m

Therefore, the jet plane travels a distance of 140 m before it stops.

Answer: 140 m

Note: The given answer range of '105,105' is incorrect as it does not fall within the calculated value range of 140 m.

Answer

Xf-xi=1/2(vi+vf)(t)

x=1/2(70+0)(3)=105

Answer

12

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