Accelleration and dispacement of planes and babiess

In summary, one type of aeroplane has a maximum acceleration of 3.5ms-2 on the ground. To reach its take-off speed of 115ms-1, it must accelerate along the runway for 82.86 seconds. The minimum length of runway needed for it to reach this speed is 4027.5 meters. Additionally, a baby carriage on a ramp accelerates at a constant rate and takes 15 seconds to reach the bottom of a 5 meter ramp. The average velocity is 3m/s, and the velocity at the bottom of the ramp is 6m/s.
  • #1
resresd
10
0
one type of aeroplane has a maximum acceleration on the ground of 3.5ms-2
a)for how many seconds must it accellerare along a runway in order to reach its take-off speed of 115ms-1?
b) what is the minimum length of runway needed for it to reach this length?

ummm, not entirely sure
av acceleration=total change in velocity/time taken

speed=distance/time

for a) i thought that it might be 115/3.5 which gives 82.86 (4 sig fig)
but then b) i thought maybe v=d/t
so 3.5=115/x
but that obviosly comes out to 115/3.5 and so the same answer, i don't really know which i have write, if either.



also i have a question about a baby carriage on a ramp
it accellerates at an average rate and takes 15s to reach the bottom of a 5m ramp.

i got that the average velocity is 15/5= 3m/s
but i need to work out the velocity at the bottom of the ramp and then the average acceleration. I can quite happily get the acceleration if i can find the final velocity, but i don't really know. i thought that as the acceleration was uniform the final velocity might be 6m/s, but this doesn't seem right
thanks
resresd
 
Physics news on Phys.org
  • #2
resresd said:
one type of aeroplane has a maximum acceleration on the ground of 3.5ms-2
a)for how many seconds must it accellerare along a runway in order to reach its take-off speed of 115ms-1?
b) what is the minimum length of runway needed for it to reach this length?

ummm, not entirely sure
av acceleration=total change in velocity/time taken

speed=distance/time

for a) i thought that it might be 115/3.5 which gives 82.86 (4 sig fig)
but then b) i thought maybe v=d/t
so 3.5=115/x
but that obviosly comes out to 115/3.5 and so the same answer, i don't really know which i have write, if either.
You considered two different ways and they come out to the same answer! Why does that bother you? In any case you shouldn't use a formula without thinking about it. Yes, "av acceleration=total change in velocity/time taken". Here you are told that the maximum acceleration is 3.5 m/s2 and that the change in velocity was 115 m/s. You are asked for the time: putting those numbers into the appropriate variables, 3.5= 115/t. Solve for t: t= 115(m/s)/3.5(m/s2)= 82.86 s. You can also check the units: to "divide" m/s by m/s2, invert the "fraction" and multiply: (m/s)(s2/m) which cancels to s (seconds) exactly what you want.


also i have a question about a baby carriage on a ramp
it accellerates at an average rate and takes 15s to reach the bottom of a 5m ramp.
I think that is intended to say it accelerates at a constant rate. "it accelerates at an average rate" doesn't make any sense.

i got that the average velocity is 15/5= 3m/s
but i need to work out the velocity at the bottom of the ramp and then the average acceleration. I can quite happily get the acceleration if i can find the final velocity, but i don't really know. i thought that as the acceleration was uniform the final velocity might be 6m/s, but this doesn't seem right
thanks
resresd
Presumably the baby carriage starts, at the top of the ramp, at 0 m/s. Now, HOW did you get "that the average velocity is 15/5= 3 m/s"? I hope you DIDN'T divide 15 s by 3 m! That would give you 15s/3m= 5 "s/m" which is NOT a unit of velocity! velocity is "distance divided by time" and you have it the wrong way around! Once you have correct average velocity you can find the velocity at the end of the ramp by using a very nice fact: as long as the acceleration is a constant, the average velocity is just the average of the velocity at the beginning and the final velocity. Since the velocity at the start is 0, the velocity at the bottom is just twice the average velocity.
 
  • #3
thank you (also I've just noticed i wrote "write" meaning "right" which is more than mildly embarraing) only for b) of the first question it was looking for a distance (i realize i may have caused confusion with my bit about the answers i got as i clearly went momentarily insane) so i don't really know what the answer would be for that as i can't now figure out what to do
 
  • #4
oh wait...v=d/t and now i have t, yet another blunder, woops! thanks
 

1. What is acceleration?

Acceleration is the rate at which the velocity of an object changes over time. It is a vector quantity, meaning it has both magnitude and direction.

2. How is acceleration calculated?

Acceleration can be calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the time interval.

3. What is displacement?

Displacement is the distance an object has moved from its starting point in a specific direction. It is also a vector quantity, meaning it has both magnitude and direction.

4. How is displacement different from distance?

Distance is the total length of the path an object has traveled, while displacement is the shortest distance between the starting and ending points of the object's motion. Distance is a scalar quantity, meaning it only has magnitude.

5. How do acceleration and displacement relate to planes and babies?

Acceleration and displacement are both important concepts in understanding the motion of planes and babies. For example, the acceleration of a plane is necessary for it to take off and maintain a steady speed, while the displacement of a baby as it crawls can help us understand its developmental milestones.

Similar threads

  • Introductory Physics Homework Help
2
Replies
56
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
958
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
11K
  • Introductory Physics Homework Help
Replies
14
Views
3K
  • Mechanics
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
2K
Back
Top