Question

A plane lands at a speed of
68 m/s and slows down at a
rate of 4 m/s^2. How much
runway needed to stop the
plane???​

Answer 1

Explanation:

u=68m/s

v=0

a=-4m/s^2 [as it is retarding acceleration]

s=?

according to third eqn

v^2 = u^2 +2as

0= (68x68) + {2x(-4s)}

0= 4624 + (-8s)

8s=4624

s= 4624/8

s=578

Therfore the distance(s)=578m

Answer 2

The correct answer is 578m.

Given: u = 68 m/s, a = -4 $\frac{m}{s^{2} }$.

To Find: How much runway needed to stop the plane.

Solution:

$v^{2}-u^{2} =2as$

v = 0 (as the plane stops)

$0^{2}-68^{2} =2*(-4)*s$

= 578 m.

Hence, the runway needed to stop the plane is 578m.

#SPJ3