What is the motion of objects under uniform acceleration?

[xCanx]

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.

 

[xCanx]

Can someone just quickly check if I am doing them right.

 

[bel]

2c)Figure out the acceleration with the information you have and apply $$s=vt-\frac{1}{2}at^2$$ with s=10.

 

[xCanx]

Thanks :) I am I doing the other ones correctly?

 

[Doc Al]

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.

 

[learningphysics]

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...

 

[xCanx]

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?

 

[xCanx]

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!

 

[Doc Al]

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?

 

[xCanx]

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

 

[Doc Al]

Doesn't say :S

You know the speed and the time!