Question

During takeoff, a plane goes from 0 to 50 m/s in 8s. What is its acceleration? Halos fast is it going after 5 s? How far had it traveled by the time it reaches 50 m/s?

Answer 1

a= 50/8 m/s^2

vf=at=50/8 * 5= 250/8 m/s at t=5sec

time to get to 50m/s
50=50/8*t or t=8 seconds
distance=1/2 a t^2=1/2 50/8 64
distance= 400 m check that.

Answer 2

Answer : Its acceleration is,
`6.25m/s^2`

The velocity after 5 s is, 31.25 m/s

The distance traveled by the time it reaches 50 m/s is, 200 m

Solution :

First we have to calculate the acceleration.

Formula used :
`a=(\Delat v)/(t)`

where,

a = acceleration

v = change in velocity = 50 - 0 = 50 m/s

t = time = 8 s

Now put all he given values in the above formula, we get

`a=(50m/s)/(8s)=6.25m/s^2`

Therefore, its acceleration is,
`6.25m/s^2`

Now we have to calculate the velocity after 5 second.

Formula used :
`v=u+at`

where,

v = final velocity

u = initial velocity = 0 m/s

a = acceleration =
`6.25m/s^2`

t = time = 5 s

Now put all the given values in the above formula, we get

`v=0m/s+(6.25m/s^2)* (5s)=31.25m/s`

Therefore, the velocity after 5 s is, 31.25 m/s

Now we have to calculate the distance traveled by the time it reaches 50 m/s.

Formula used :
`v^2-u^2=2as`

where,

s = distance traveled

Now put all the given values in the above formula, we get

`(50m/s)^2-(0m/s)^2=2* (6.25m/s^2)* s`

`s=200m`

Therefore, the distance traveled by the time it reaches 50 m/s is, 200 m