Relative Velocity of Car B to Car A at Different Speeds

[missie]

Question: Look at figure 1.6. if Car a slowed to 30m/s. what would the velocity of car b relative to car a be?

figure 1.6 is a picture of 2 car going forward the same way. Car a is in the front and car b is in the back. Car a is going at 40m/s and car b is going 60 m/s.

will this be how i solve it?

40m/s - 60m/s = -20 m/s

is that how the answer is or I'm i wrong? is it another way?

 

[LowlyPion]

Question: Look at figure 1.6. if Car a slowed to 30m/s. what would the velocity of car b relative to car a be?

figure 1.6 is a picture of 2 car going forward the same way. Car a is in the front and car b is in the back. Car a is going at 40m/s and car b is going 60 m/s.

will this be how i solve it?

40m/s - 60m/s = -20 m/s

is that how the answer is or I'm i wrong? is it another way?

But was that the actual question?

 

[missie]

yes that was the question.

 

[missie]

the problem is that i don't know if i solved it right.

 

[LowlyPion]

yes that was the question.

Please read the problem more carefully.

Look at figure 1.6. if Car a slowed to 30m/s. what would the velocity of car b relative to car a be?

 

[missie]

so the answer should be:

30m/s - 60m/s = -30m/s

? and thanks ;]

 

[LowlyPion]

so the answer should be:

30m/s - 60m/s = -30m/s

? and thanks ;]

Not quite.

Take the vector difference by subtracting the vector of car A from the vector of Car B.

In the case where they were moving away the distance was getting bigger. Subtracting from the second really resulted in adding. In this case the vector of the second is larger and when you subtract the vector (of the one you want it relative to) it's not (-) it's (+). Reverse the order of the subtraction in the equation you gave.

 

[missie]

thanks so much ;]