Question

a driver of a car traveling at 15.0 m/s applies the breaks, causing a uniform acceleration of -2.0 m/s^2. how long does it take the car to accelerate to a final speed of 10.0 m/s?
i just need to know the delta x, delta time and vi and the vf please!!!

Answer 1

vi is initial velocity.
It began at 15.0 m/s as you can see from your question.
vf is final velocity.
The driver is decelerating (slowing down) to 10 m/s.

Answer 2

Delta x is a change in position, so it would be the distance.

Answer 3

You can use V = U + at to find the time required.

Answer 4

V = Final velocity
U = initial velocity
a = retardation/deceleration
t = time taken

Answer 5

\[\Delta x=x _{f}-x _{i}\]
\[x _{f}=x _{0}+v _{i}t+\frac{ 1 }{ 2 }at^2\]
\[x _{f}-x _{i}=v _{i}t+\frac{ 1 }{ 2 }at^2\]
\[\Delta x=v _{i}t+\frac{ 1 }{ 2 }at^2\]

Answer 6

Do
\[v _{f}=v _{i}+at\]
first to get time first.

Answer 7

Ramanujan!!

Answer 8

Nikola Tesla !!

Answer 9

I am going to try an make sure that your understanding of this topic is clear;
So someone is driving a car with a speed of 15 m/s. This is the velocity that the question starts with making it the 'initial' velocity. acceleration is a change in speed, if it is negative, the speed is going down, if not then the speed is increasing. So if someone is decelerating at a rate of -2m/s^2 then that means that every second, his speed is going down by 2m/s.
Example: 15m/s is the 'initial' velocity, after one second the velocity will decrease by 2m/s which will give us 15-2=13m/s. Now we already know how slow the diver wants to go which is 10m/s from 15m/s. This means the driver wants to decrease his speed by 5m/s at a rate of -2m/s^2, so after 2 seconds he will have slowed down by 4m/s to give us 11m/s and after another half second he will have reached 10m/s which is his final velocity, and he took 2.5 seconds to do that.

If you still can't do this problem 'without thinking' please tell me what you haven't understood.