I believe it should rather be 'accelerates at -2m/s2'. Negative acceleration is deceleration and negative deceleration is acceleration..Grace204 said:decelerates at -2.00 m/s
I don't know how to. The anwer I am getting is -18.75, and i don't think that's right.cnh1995 said:Your relevant equation is correct. Put the known values in that equation and solve for displacement.
s=(v2-u2)/2a
It is negative because you used a=2m/s2Grace204 said:I don't know how to. The anwer I am getting is -18.75, and i don't think that's right.
Grace204 said:This is what i did-
(5.00 m/s [W])2 - (10.0 m/s [W])2 / 2 * -(-2.00 m/s2
(-5.00 m/s)2 - (-10.0 m/s)2 / 4.00 m/s2
After this, what happens to the units? How do the s2 cancel each other out?
Yes. And only the unit of displacement remains which is meter..Every equation has to be dimensionally correct.Grace204 said:This is what i did-
(5.00 m/s [W])2 - (10.0 m/s [W])2 / 2 * -(-2.00 m/s2
(-5.00 m/s)2 - (-10.0 m/s)2 / 4.00 m/s2
After this, what happens to the units? How do the s2 cancel each other out?
So s2 divided by s2 cancels each other out and does not equal s4? Because acceleration is m/s divided by scnh1995 said:Yes. And only the unit of displacement remains which is meter..Every equation has to be dimensionally correct.
Why did you put double negative signs on the acceleration in the denominator?Grace204 said:This is what i did-
(5.00 m/s [W])2 - (10.0 m/s [W])2 / 2 * -(-2.00 m/s2
(-5.00 m/s)2 - (-10.0 m/s)2 / 4.00 m/s2
After this, what happens to the units? How do the s2 cancel each other out?
Yes. You can see s2 cancel each other and m2/m becomes m.Grace204 said:m2/s2 divided by m/s2?
SteamKing said:Why did you put double negative signs on the acceleration in the denominator?
You should be able to work out the resulting units using standard algebra.
If you square velocity in m/s, what do you get?
If you divide these units for velocity squared by the units for acceleration, what's left?
Is this wrong? If so, how would I write this out?Grace204 said:I put double negatives on the denominator because the question says is "decelerates at -2.00 m/s2 [W] or -(-2.00 m/s2)"
The question is poorly worded. One negative sign should suffice, since 'decelerate' means to reduce speed or velocity.Grace204 said:Is this wrong? If so, how would I write this out?
The initial and final velocities are 10m/s and 5m/s respectively. This clearly tells you that the car is 'decelerating'. So, you should write either 'a car is decelerating at 2m/s2' or ' a car is accelerating at -2m/s2'. That's what I meant in #2.Grace204 said:Is this wrong? If so, how would I write this out?
Now substitute a=-2 here, since the car is decelerating.cnh1995 said:s=(v2-u2)/2a
Displacement is a measure of the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction, and is typically measured in units of length, such as meters or feet.
Displacement can be calculated by multiplying an object's average velocity by the time it takes to travel that distance. This can be represented mathematically as Δx = v * t, where Δx is displacement, v is velocity, and t is time.
Velocity and displacement are closely related, as displacement is defined as the change in an object's position over time, and velocity is the rate of change of an object's displacement over time. In other words, velocity is the derivative of displacement with respect to time.
Deceleration, also known as negative acceleration, is when an object's velocity decreases over time. This means that the object is traveling in the opposite direction of its initial velocity. Deceleration can cause a decrease in displacement, as the object is moving in the opposite direction of its initial displacement.
Yes, displacement can be negative. This means that the object has moved in the opposite direction from its starting point. For example, if an object starts at a position of 5 meters and ends at a position of 2 meters, its displacement would be -3 meters. This indicates that the object moved 3 meters in the negative direction from its starting point.