Question

A plane has take of speed of 88.3 m/s and requires 1.365 m to reach that speed determine the acceleration of the of yhe plane and the time required to reach this speed

Answer 1

Answer:

s=1.365 u=0 v=88.3 a=?

v^2-u^2=2as

88.3^2 - 0^2=2* a *1.365

7796.89=2.73*a

7796.89/2.73=a

a=2856.003663m/s^2

Explanation:

i used kinematics formula 3 and solved this question

hope this helps

Answer 2

Correct Question-

A plane has take of speed of 88.3 m/s and requires 1365 m to reach that speed determine the acceleration of the of the plane and the time required to reach this speed.

Solution-

Given that, a plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed.

From the above data, we have v is 88.3 m/s, u is 0 m/s and s is 1.365 m.

We have to find the acceleration of the of the plane and the time required to reach this speed.

For acceleration:

Using the Third Equation Of Motion,

v² - u² = 2as

Substitute the known values,

(88.3)² - (0)² = 2a(1365)

7796.89 - 0 = 2730a

7796.89/2730 = a

2.856 = a

3 = a

Therefore, the acceleration of the plane is 3 m/s².

For time:

Using the First Equation Of Motion,

v = u + at

Substitute the values,

88.3 = 0 + 3t

88.3/3 = t

29.43 = t

Therefore, the time taken by the plane is 29.43 sec.