The driver of a car slams on the brakes, causing the car to slow down at a rate of
20.0ft/s as the car skids 295ft to a stop.
How long does the car take to stop?
What was the car’s initial speed?

The Driver Of A Car Slams On The Brakes, Causing The Car To Slow Down At A Rate Of20.0ft/s As The Car

Answers

Answer 1

Answer:

The time taken for the car to stop is 5.43 s.

The initial velocity of the car is 108.6 ft/s

Explanation:

The following data were obtained from the question:

Acceleration (a) = –20 ft/s² (since the car is coming to rest)

Distance travalled (s) = 295 ft

Final velocity (v) = 0 ft/s

Time taken (t) =?

Initial velocity (u) =?

Next, we shall determine the initial velocity of the car as shown below:

v² = u² + 2as

0² = u² + (2 × –20 × 295)

0 = u² + (–11800)

0 = u² – 11800

Collect like terms

0 + 11800 = u²

11800 = u²

Take the square root of both side

u = √11800

u = 108.6 ft/s

Therefore, the initial velocity of the car is 108.6 ft/s.

Finally, we shall determine the time taken for the car to stop as shown below:

Acceleration (a) = –20 ft/s² (since the car is coming to rest)

Final velocity (v) = 0 ft/s

Initial velocity (u) = 108.6 ft/s

Time taken (t) =?

v = u + at

0 = 108.6 + (–20 × t)

0 = 108.6 + (–20t)

0 = 108.6 – 20t

Collect like terms

0 – 108.6 = – 20t

– 108.6 = – 20t

Divide both side by –20

t = – 108.6 / –20

t = 5.43 s

Therefore, the time taken for the car to stop is 5.43 s.


Related Questions

A rock is thrown downward from the top of a cliff with an initial speed of -13m/s.If the rock hits the ground after 2.7 s, what is the height of the cliff

Answers

Answer:

0.62 m down.

Explanation:

Use the formula d= vi*t +1/2 at^2

So the initial speed or vi is -13 m/s and time is 2.7 s and a or acceleration is 9.8 m/s^2 down ( because it's the gravity). So you plugged all the values and you will get 0.62 m. We put the down because the acceleration is saying down. Because displacement is vector quantity you have to show the direction of it.

Choose Yes or No to tell if the number 9.81 will make each equation true.



12.47 – □ = 2.66

8.05 + □ = 17.96

14.08 – □ = 4.17

3.67 + □ = 13.48

Answers

Answer:

yes

no

no

yes

Explanation:

An 8000.0 kg truck starts off from rest and reaches a velocity of 18.0 m/s in 6.00 seconds. What is the truck’s acceleration and how much momentum does it have after it has reached this final velocity?

Answers

Explanation:

Acceleration is change in velocity over time.

a = Δv / Δt

a = (18.0 m/s − 0 m/s) / (6.00 s)

a = 3.00 m/s²

Momentum is mass time velocity.

p = mv

p = (8000 kg) (18.0 m/s)

p = 144,000 kg m/s

When an 8000.0 kg truck starts off from rest and reaches a velocity of 18.0 m/s in 6.00 seconds then the truck’s acceleration would be 3 m/s² and the momentum after it has reached this final velocity would be 144000 kgm/s.

What is momentum?

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

Mathematically the formula of the momentum is

P = mv

where P is the momentum of the particle

m is the mass of the particle

v is the velocity by which the particle is moving

As given in the problem An 8000.0 kg truck starts off from rest and reaches a velocity of 18.0 m/s in 6.00 seconds

By using the first equation is given as follows

v = u + at

18 =0 + a×6

a = 3 m/s²

As given in the problem the final velocity of An 8000.0 kg truck is 18 m/s

the momentum is given by the formula

P = mv

P= 8000×18

P = 144000 kgm/s

Thus, the truck’s acceleration would be 3 m/s², and the momentum after it has reached this final velocity would be 144000 kgm/s.

Learn more about momentum from here

brainly.com/question/17662202

#SPJ5

Which of the following statements are true

Answers

Answer:

none of them

Explanation:

Answer:

Did u forget to add a pic?

Explanation:

What type of force occurs when one object pushes or pulls a second object while they
are touching?

Answers

★・・・・・・★・・・・・・★・・・・・・★・・・・・・★・・・・・・

»Answer

【Contact forces involve push, pull and friction. A contact force is when two interacting objects are touching, for example: when you throw a ball you are using a contact force.】

hope it helps

can u give a thanks

thank you

★・・・・・・★・・・・・・★・・・・・・★・・・・・・★・・・・・・

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