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Anton [14]
3 years ago
6

Two cars, A and B , travel in a straight line. The distance of A from the starting point is given as a function of time by xA(t)

=αt+βt2, with α=2.60m/s and β=1.20m/s2. The distance of B from the starting point is xB(t)=γt2−δt3, with γ=2.80m/s2 and δ=0.20m/s3
Part A. Which car is ahead just after they leave the starting point?
car A
car B
Part B. At what time(s) are the cars at the same point?
Express your answer numerically. If there is more than one answer, enter each answer separated by a commas.
Part C. At what time(s) is the distance from A to B neither increasing nor decreasing?
Express your answer numerically. If there is more than one answer, enter each answer separated by a comma.
Part D. At what time(s) do A and B have the same acceleration?
Express your answer numerically. If there is more than one answer, enter each answer separated by a comma.
Physics
1 answer:
OLga [1]3 years ago
3 0

A) Car A is initially ahead

B) The two cars are at the same point at the times: t = 0, t = 2.27 s and

t = 5.73 s

C) The distance between the two cars is not changing at t = 1.00 s and t = 4.33 s

D) The two cars have same acceleration at t = 2.67 s

Explanation:

A)

The position of the two cars at time t is given by the following functions:

x_A(t) = \alpha t + \beta t^2

with

\alpha = 2.60 m/s\\\beta = 1.20 m/s^2

Substituting,

x_A(t)=2.60t+1.20 t^2

And

x_B(t)=\gamma t^2 - \delta t^3

with

\gamma=2.80 m/s^2\\\delta = 0.20 m/s^3

Substituting,

x_B(t)=2.80t^2-0.20t^3

Here we want to find which car is ahead just after they leave the starting point. To find that, we just need to calculate the position of the two cars after a very short amount of time, let's say at t = 0.1 s. Substituting this value into the two equations, we get:

x_A(0.1)=2.60(0.1)+1.20(0.1)^2=0.27 m

x_B(0.1)=2.80(0.1)^2-0.20(0.1)^3=0.03 m

So, car A is initially ahead.

B)

The two cars are at the same point when their position is the same. Therefore, when

x_A(t)=x_B(t)

which means when

2.60t+1.20t^2 = 2.80t^2-0.20t^3

Re-arranging the equation, we find

0.20t^3-1.6t^2+2.60t=0\\t(0.20t^2-1.6t+2.60)=0

One solution of this equation is t = 0 (initial point), while we have two more solutions given by the equation

0.20t^2-1.6t+2.60=0

which has two solutions:

t = 2.27 s

t = 5.73 s

So, these are the times at which the cars are at the  same point.

C)

The distance between the two cars A and B is not changing when the velocities of the two cars is the same.

The velocity of car A is given by the derivative of the position of  car A:

v_A(t) = x_A'(t)=(2.60t+1.20t^2)'=2.60+2.40t

The velocity of car B is given by the derivative of the position of car B:

v_B(t)=x_B'(t)=(2.80t^2-0.20t^3)'=5.60t-0.60t^2

Therefore, the distance between the two cars is not changing when the two velocities are equal:

v_A(t)=v_B(t)\\2.60+2.40t=5.60t-0.60t^2\\0.60t^2-3.20t+2.60=0

This is another second-order equation, which has two solutions:

t = 1.00 s

t = 4.33 s

D)

The acceleration of each car is given by the  derivative of the velocity of the car A.

The acceleration of car A is:

a_A(t)=v_A'(t)=(2.60+2.40t)'=2.40

While the acceleration of car B is:

a_B(t)=v_B'(t)=(5.60t-0.60t^2)'=5.60-1.20t

So, the two cars have same acceleration when

a_A(t)=a_B(t)

And solving the equation, we find:

2.40=5.60-1.20t\\1.20t=3.20\\t=2.67 s

So, the two cars have same acceleration at t = 2.67 s.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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