Uniform acceleration problems

  • Thread starter xCanx
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  • #1
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I'm having trouble with two problems:

1. A driver is travelling 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.
 

Answers and Replies

  • #2
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Can someone just quickly check if Im doing them right.
 
  • #3
bel
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2c)Figure out the acceleration with the information you have and apply [tex]s=vt-\frac{1}{2}at^2[/tex] with s=10.
 
  • #4
45
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Thanks :) I am I doing the other ones correctly?
 
  • #5
Doc Al
Mentor
45,248
1,598
1. A driver is travelling 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.
 
  • #6
learningphysics
Homework Helper
4,099
6
I'm having trouble with two problems:

1. A driver is travelling 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...
 
  • #7
45
0
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?
 
  • #8
45
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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!
 
  • #9
Doc Al
Mentor
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1,598
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
45
0
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
Doc Al
Mentor
45,248
1,598
Doesn't say :S
You know the speed and the time!
 

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