An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.
Question
An airplane accelerates down a runway at 3.20 m/s² for 32.8 s until it finally lifts off the ground. Determine the distance traveled before takeoff.
Solution
To solve this problem, we can use the formula for distance in uniformly accelerated motion:
d = vit + 0.5at^2
where:
- d is the distance,
- vi is the initial velocity,
- a is the acceleration, and
- t is the time.
In this case, the airplane starts from rest, so the initial velocity (vi) is 0. The acceleration (a) is 3.20 m/s^2 and the time (t) is 32.8 s. Substituting these values into the formula gives:
d = 0*(32.8 s) + 0.5*(3.20 m/s^2)*(32.8 s)^2
Solving this equation gives the distance traveled before takeoff.
Similar Questions
An airplane accelerates down a runway at 4 m/s2 for 24 s until it finally lifts off the ground. Determine the distance traveled before takeoff.
An airplane from rest accelerates on a runway at 5.50 m/s2 for 20.25 s until it finallytakes off the ground. What is the distance covered before takeoff?
A car begins decelerating at 2 m/s2 and takes 30 meters to stop. How long did the car take to stop?≈ 5.5 s≈ 7.5 s ≈ 4.5 s≈ 8.5 s
A car accelerates from 10.0 m/s to 30.0 m/s at a rate of 3.00 m/s2.How far does the car travel while accelerating?Express your answer in meters.
What is the velocity of an object accelerating at −25.0m/s 2 after 3.0s when it is initially moving at 30.0m/s?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.