What is the motion of objects under uniform acceleration?

AI Thread Summary
The discussion revolves around two physics problems involving uniform acceleration. In the first scenario, a driver traveling at 25 m/s takes 1 second to react before braking at a rate of 3.0 m/s², ultimately stopping 5 meters short of a bridge. The time required to stop after braking is calculated to be approximately 8.3 seconds, and the total distance from the bridge when the sign was first seen is determined to be 109 meters, accounting for the reaction time. The second problem involves a box that falls and slides 45 meters before stopping, with the time taken to slide calculated as 6 seconds. Participants emphasize the importance of including the initial reaction time in the driver's stopping distance and clarify the need to find the acceleration of the box for further calculations.
xCanx
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I'm having trouble with two problems:

1. A driver is traveling at 25 m/s when she spots a sign that reads "BRIDGE OUT AHEAD". It takes her 1.0 s to react and begin braking. The car slows down at a rate of 3.0 m/s^2.
Luckily, she stops 5.0 m short of the washed out bridge.

a) How much time was required to stop the car once the brakes were applied?

My attempt:

t= 0 m/s - 25 m/s (divided by) 3.0 m/s^2
t= -8.3
t =(8.3 s)


b) How far was the driver from the bridge when she first noticed the sign?

My attempt:

d = 0.5(V1 + V2)t
d = 0.5(0 + 25) 8.3
d = 104 + 5.0 (she fell short)
d = 109 m


2. A box accidently falls from the back of a truck and hits the ground with a speed of
15 m/s. it slides along the ground for a distance of 45 m before coming to rest.

Determine:

a) the length of time the box slides before stopping

My attempt:

d = 0.5(V1 + V2)t
45 = 0.5(15 + 0) t
solve for t and get
t = 6


c) the time it takes to slide the last 10 m

I don't know this one.
 
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Can someone just quickly check if I am doing them right.
 
2c)Figure out the acceleration with the information you have and apply s=vt-\frac{1}{2}at^2 with s=10.
 
Thanks :) I am I doing the other ones correctly?
 
xCanx said:
1. A driver is traveling at 25 m/s when she spots a sign that reads "BRIDGE OUT AHEAD". It takes her 1.0 s to react and begin braking. The car slows down at a rate of 3.0 m/s^2.
Luckily, she stops 5.0 m short of the washed out bridge.

a) How much time was required to stop the car once the brakes were applied?

My attempt:

t= 0 m/s - 25 m/s (divided by) 3.0 m/s^2
t= -8.3
t =(8.3 s)


b) How far was the driver from the bridge when she first noticed the sign?

My attempt:

d = 0.5(V1 + V2)t
d = 0.5(0 + 25) 8.3
d = 104 + 5.0 (she fell short)
d = 109 m
Don't forget that it takes her 1.0 second to react.
2. A box accidently falls from the back of a truck and hits the ground with a speed of
15 m/s. it slides along the ground for a distance of 45 m before coming to rest.

Determine:

a) the length of time the box slides before stopping

My attempt:

d = 0.5(V1 + V2)t
45 = 0.5(15 + 0) t
solve for t and get
t = 6


c) the time it takes to slide the last 10 m
Start by finding the acceleration.
 
xCanx said:
I'm having trouble with two problems:

1. A driver is traveling at 25 m/s when she spots a sign that reads "BRIDGE OUT AHEAD". It takes her 1.0 s to react and begin braking. The car slows down at a rate of 3.0 m/s^2.
Luckily, she stops 5.0 m short of the washed out bridge.

a) How much time was required to stop the car once the brakes were applied?

My attempt:

t= 0 m/s - 25 m/s (divided by) 3.0 m/s^2
t= -8.3
t =(8.3 s)

I'd do it like this (0-25)/(-3)... since it is slowing down acceleration is -3...

b) How far was the driver from the bridge when she first noticed the sign?

My attempt:

d = 0.5(V1 + V2)t
d = 0.5(0 + 25) 8.3
d = 104 + 5.0 (she fell short)
d = 109 m

You also have the 1.0s before she reacted... so you need to add 25m/s*1.0s = 25m

2. A box accidently falls from the back of a truck and hits the ground with a speed of
15 m/s. it slides along the ground for a distance of 45 m before coming to rest.

Determine:

a) the length of time the box slides before stopping

My attempt:

d = 0.5(V1 + V2)t
45 = 0.5(15 + 0) t
solve for t and get
t = 6
Looks good.

c) the time it takes to slide the last 10 m

I don't know this one.

find the acceleration of the block...
 
Doc Al said:
Don't forget that it takes her 1.0 second to react.

Start by finding the acceleration.

So I would subtract 1.0 s from 8.3?
 
learningphysics said:
I'd do it like this (0-25)/(-3)... since it is slowing down acceleration is -3...



You also have the 1.0s before she reacted... so you need to add 25m/s*1.0s = 25m


Looks good.



find the acceleration of the block...
Thanks :) That helped a lot!
 
xCanx said:
So I would subtract 1.0 s from 8.3?
No. The 8.3s is for the accelerated portion, which you've already figured out. How far does she go before she even hits the brakes?
 
  • #10
Doc Al said:
No. The 8.3s is for the accelerated portion, which you've already figured out. How far does she go before she even hits the brakes?
Doesn't say :S
 
  • #11
xCanx said:
Doesn't say :S
You know the speed and the time!
 
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