Question

A jet lands on an aircraft
carrier at 63 m/s. What is its
acceleration in m/s2 if it stops
in 2.0 s?*​

Answer 1

Given:-

• Initial velocity (u) = 63m/s

• Final Velocity (v) = 0m/s

• Time taken (t) = 2 s

To Find:-

• Acceleration of jet (a).

Solution:-

As we know that the rate of change of velocity at per unit time is called acceleration.

→ a = v-u/t

Substitute the value we get

→ a = 0-63/2

→ a = -63/2

→ a = -31.5m/s²

Negative sign show retardation

∴ The acceleration of the jet is 31.5m/s²

Answer 2

Given :-

• Initial velocity (u) = 63m/s.
• Final velocity (v) = 0m/s.
• Time = 2.0sec

To Find :-

• What is acceleration(a) of jet.

Solution :-

☀ Finding acceleration of jet. ☀

↪Applying 1st equation of motion, We get :

↪v = u + at

↪ 0 = 63 + a × 2.0

↪ -63 = 2.0a

↪a = -63/2.0

↪ a = -31.5m/s².

[ Negative (-) Sign show Retardation ]

↪a = 31.5 m/s².

.°. Acceleration is 31.5m/s².