Respuesta :
The question is incomplete. Here is the complete question.
Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance [tex]D_{A}[/tex] beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed [tex]v_{A}[/tex]. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed [tex]v_{B}[/tex], which is greater than [tex]v_{A}[/tex].
Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.
Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.
Answer: Part A: [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]
Part B: [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]
Explanation: First, let's write an equation of motion for each car.
Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:
[tex]x=x_{0}+vt[/tex]
where
[tex]x_{0}[/tex] is initial position
v is velocity
t is time
Car A started the race at a distance. So at t = 0, initial position is [tex]D_{A}[/tex].
The equation will be:
[tex]x_{A}=D_{A}+v_{A}t[/tex]
Car B started at the starting line. So, its equation is
[tex]x_{B}=v_{B}t[/tex]
Part A: When they meet, both car are at "the same position":
[tex]D_{A}+v_{A}t=v_{B}t[/tex]
[tex]v_{B}t-v_{A}t=D_{A}[/tex]
[tex]t(v_{B}-v_{A})=D_{A}[/tex]
[tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]
Car B meet with Car A after [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex] units of time.
Part B: With the meeting time, we can determine the position they will be:
[tex]x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )[/tex]
[tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]
Since Car B started at the starting line, the distance Car B will be when it passes Car A is [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex] units of distance.
The distance traveled by the car A and car B should be equal to the as they meet at the same position.
The time car B will catch the car A after is,
[tex]\dfrac{D_A}{v_B-v_A}[/tex]
How to calculate the distance traveled by body?
The distance is the product of the speed of the body and the time taken to travel the distance.
Given information-
Car A has a head start and is a distance DA beyond the starting line at,
[tex]t=0[/tex]
Car A travels at a constant speed [tex]v_A[/tex].
Car B travels at a constant speed [tex]v_B[/tex].
The distance is the product of the speed of the body and the time taken to travel the distance.
The position equation from the motion for car A can be given as,
[tex]x_A=v_At+D_A[/tex]
The position equation from the motion for car B can be given as,
[tex]x_B=v_Bt[/tex]
The distance traveled by the car A and car B should be equal to the as they meet at the same position. Thus,
[tex]x_A=x_B[/tex]
Put the values,
[tex]v_At+D_A=v_Bt\\v_At-v_Bt=-D_A\\t(v_B-v_A)=D_A\\t=\dfrac{D_A}{v_B-v_A}[/tex]
Hence the time car B will catch the car A after is,
[tex]\dfrac{D_A}{v_B-v_A}[/tex]
Learn more about the speed of the object here;
https://brainly.com/question/4931057
