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maw [93]
3 years ago
10

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

e DA beyond the starting line at t=0. The starting line is at x=0. Car A travels at a constant speed vA. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed vB, which is greater than vA
How long after Car B started the race will Car B catch up with Car A?
Physics
1 answer:
kumpel [21]3 years ago
7 0

Answer:\frac{D_A}{v_B-v_A}

Explanation:

Given

car A had a head start of D_A

and it starts at x=0 and t=0

Car B has to travel a distance of D_A and d_a

where d_a is the distance travel by car A in time t

distance travel by car A is

d_a=v_A\times t

For car B with  speed v_B

d_B=D_A+d_a

v_B\times t=D_A+v_A\times t

t=\frac{D_A}{v_B-v_A}

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The universe tends to distribute energy evenly; that is, to maximize entropy. Intuitively, entropy is a physical quantity that, by calculation, allows us to determine the part of energy per unit of temperature that cannot be used to produce work.

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2 years ago
Ls -2 a solution of 4x +3= -5?.<br>​
torisob [31]

Answer:

yes

Explanation:

Let's solve your equation step-by-step.

4x+3=−5

Step 1: Subtract 3 from both sides.

4x+3−3=−5−3

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Step 2: Divide both sides by 4.

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Hope it helps,

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