Respuesta :
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
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
∴ s = 0 + 0.5 × 4× 40²
∴ s = 3200 meter
The velocity of the plane at takeoff and distance traveled by plane is 160 m/s and 3200 meters respectively.
What is the equation of motion?
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The first equation of motion is given as,
[tex]v=u+at[/tex]
The second equation of the motion for distance can be given as,
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Here, (i) is the initial speed of the body, (a) is the acceleration of the body and (t) is the time taken by it.
The acceleration of the plane at the time of takeoff is 4 m/s² and the time taken by the plane to reach the takeoff speed is 40 seconds.
- a) The velocity of the plane at takeoff-
As the initial velocity of the plane is zero. Thus the velocity of the plane at takeoff can be given using the first equation of motion as,
[tex]v=0+4\times40\\v=\rm 160m/s[/tex]
Thus, the velocity of the plane at takeoff is 160 m/s.
- b) The distance did the plane travel in that time-
As the initial velocity of the plane is zero. Thus the distance the plane travel in 40 seconds after takeoff can be given using the second equation of motion as,
[tex]s=0+\dfrac{1}{2}4\times40^2\\s=3200\rm m[/tex]
The distance did the plane travel in that time is 3200 meters.
Hence, the velocity of the plane at takeoff and distance traveled by plane is 160 m/s and 3200 meters respectively.
Learn more about the equation of motion here;
https://brainly.com/question/13763238
