Question

A jet plane beginning to take off moves down the runaway with a constant acceleration of 4m/s2.Find: (a)Position and velocity after 5 secs (b) If a speed of 70m/sec is required for the plane to leave the ground, How long the runaway is required?
Wrong ans would be reported

Answer 1

Answer:

Given a = 4 m/s^2

u = 0

v= u + at

a) at t = 5 s

position s = ut + 1/2at^2

s= 0 + 1/2 x 4 x 5^2

s = 50 m

v = u+ at

= 0 + 4x5

= 20 m/s

b) v= 70 m/s

t= v/a = 70/4 = 17.5 s

s= ut + 1/2 at^2

t = 17.5s ; a = 4 ; u = 0

Length of runway s= 0 + 1/2 x 4 x 17.5 x 17.5

= 2 x 17. x 17.5 m

= 2 x 306.25 m

= 612.5 m

Answer 2

(a) After 5 seconds the position of jet is 50 meters and velocity is 20 m/s

(a) After 5 seconds the position of jet is 50 meters and velocity is 20 m/s (b) Length of runaway is 612.5 meter.

GIVEN

Acceleration = 4m/s²

TO FIND

Position and velocity after 5 seconds

Length of the runaway.

SOLUTION

We can simply solve the above problem

Acceleration = 4 m/s²

Initial velocity = 0

Time = 5 seconds

Position after 5 seconds

$s \: = ut \: + \frac{1}{2} a {t}^{2}$

$= \frac{1}{2} \times 4 \times 5 \times 5$

= 50 m

v = u + at

= 4 × 5

= 20 m/s

(b) Final velocity = 70 m/s

t = v/a

= 70/4

= 17.5 seconds

Length of the runaway,

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

$= 2 \times 17.5 \times 17.5$

= 612.5 meter.

Hence,

Hence, (a) After 5 seconds the position of jet is 50 meters and velocity is 20 m/s

Hence, (a) After 5 seconds the position of jet is 50 meters and velocity is 20 m/s (b) Length of runaway is 612.5 meter.

Hence, (a) After 5 seconds the position of jet is 50 meters and velocity is 20 m/s (b) Length of runaway is 612.5 meter. #Spj2