Question

A plane has a take off speed of 1 point

300 km/h. What is the
acceleration in m/s2 of the
plane if the plane started from
rest and took 45 seconds to
take off? *​

Answer 1

Acceleration = 1.85m/s²

Step-by-step explanation:

Given the following data;

Final velocity = 300km/h to m/s = 300*1000/3600 = 83.33m/s

Time = 45 seconds

Since the plane started from rest, initial velocity is equal to 0m/s.

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Mathematically, acceleration is given by the equation;

`Acceleration (a) = (final \; velocity - initial \; velocity)/(time)`

Where,

a is acceleration measured in
`ms^(-2)`

v and u is final and initial velocity respectively, measured in
`ms^(-1)`

t is time measured in seconds.

Substituting into the equation, we have;

Acceleration = (83.33 - 0)/45

Acceleration = 83.33/45

Acceleration = 1.85m/s²

Therefore, the acceleration of the plane is 1.85m/s².