Constant speed and deaccelaration

In summary, the conversation discusses a problem involving a car traveling at a constant speed of 18 m/s and suddenly braking. The question asks for the distance traveled before braking and the distance traveled before coming to a complete stop. The equations needed to solve the problem are mentioned, including the formula for constant speed and the formula for constant acceleration. The correct values and units are also clarified. Finally, the conversation ends with a humorous comment about the difference between "breaks" and "brakes".
  • #1
access
18
0

Homework Statement


R is traveling in a car at a constant speed of 18ms when she suddenly breaks.

if reaction time is .52s for applying foot to break how far is traveled before break is applied?

and if the break cause a deceleration of 5m/ssquared, how is traveled before stopping?
 
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  • #2
We won't do your homework for you. You have to show at least an attempt at a solution, and explain where you're stuck.
 
  • #3
ok sorry i not sure on the equations needed i have read the textbook. can you give me the equations so i can work them out? i thought mayb speed / time would give distance?
 
  • #4
is it constant speed times time? if i do this tho my equation is 18 X .52 yes? 9.36 metres?
 
  • #5
then the distance of deeceration is constant speed / 5mssquared?
 
  • #6
sorry i don't mean 5mssquared i ment 5 ms per s
 
  • #7
The constant speed of the car is 18 m/s and not 18 ms.

In case of constant speed the distance traveled is speed times time. Yes, the car traveled 9.36 meters before braking.
Constant acceleration means that the change of velocity is proportional to the time. (v2-v1)=at. In case of deceleration, v2<v1 and a is negative. You can get t from the change of velocity and a.
In case of accelerating motion, the displacement is
x= v0*t +a/2 *t^2.
Plug in the stopping time for t, and the appropriate value for a.

ehild
 
  • #8
thankyou but this is to work out the distance yeah? v0=constant velocity? time taken to stop would be 18/5? so t = 3.6 so (18*3.6) + (5/2*3.6^2) ?
 
  • #9
access said:
thankyou but this is to work out the distance yeah? v0=constant velocity? time taken to stop would be 18/5? so t = 3.6 so (18*3.6) + (5/2*3.6^2) ?

It is deceleration, so you need to plug in a=-5.

ehild
 
  • #10
props to you.
 
  • #11
When the car breaks, it goes to pieces. When the car brakes, it slows and stops.
 
  • #12
SteamKing said:
When the car breaks, it goes to pieces. When the car brakes, it slows and stops.
Not the car breaks, but the lady who drives it. :smile:

ehild
 

FAQ: Constant speed and deaccelaration

What is constant speed and deacceleration?

Constant speed and deacceleration refers to the motion of an object that maintains a steady velocity while also slowing down at a constant rate.

How is constant speed and deacceleration different from constant speed and acceleration?

In constant speed and acceleration, the object maintains a steady velocity while also increasing in speed at a constant rate. In constant speed and deacceleration, the object maintains a steady velocity while also decreasing in speed at a constant rate.

What is the formula for calculating deacceleration?

The formula for deacceleration, also known as deceleration, is: a = (v - u)/t, where:a = deacceleration (m/s^2)v = final velocity (m/s)u = initial velocity (m/s)t = time (s)

What are some examples of objects experiencing constant speed and deacceleration?

Examples of objects experiencing constant speed and deacceleration include a car that is gradually braking to come to a stop, a ball rolling up a hill and slowing down, and a rocket that is gradually slowing down as it enters the Earth's atmosphere.

How is deacceleration measured and what are some common units?

Deacceleration is typically measured in meters per second squared (m/s^2) or in units of gravity (g). Some common units for deacceleration are meters per second squared, kilometers per hour squared, and feet per second squared.

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