Question

Plz help..During takeoff a plane accelerates at 4m/s^2 and takes 40s to reach takeoff speed.

Answer 1

Answer:

The velocity of the plane at take off is 160 m/s.

The distance travel by the plane in that time is 3200 meter.

Explanation:

Given:

Acceleration, a = 4 m/s²

Time, t = 40 s

u = 0 i .e initial velocity

To Find:

velocity , v = ?

distance , s =?

Solution:

we have first Kinematic equation

v = u + at

∴ v = 0 + 4×40

∴ v = 160 m/s

Now by Third Kinematic equation

$s = ut + \frac{1}{2}at^{2}$

∴ s = 0 + 0.5 × 4× 40²

∴ s = 3200 meter