Respuesta :
Answer:
Car B reaches car A in 19.7 s.
Explanation:
Hi there!
The equation of the position of an object moving in a straight line at constant acceleration is as follows:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the object at time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration
When both cars meet, their positions are the same. At the meeting point:
position of car A = position of car B
xA = xB
x0A + v0A · t + 1/2 · aA · t² = x0B + v0B · t + 1/2 · aB · t²
Let´s place the origin of the frame of reference at the point where A is located. In that case x0A = 0 and x0B = 2900 m. Since both cars are initially at rest, v0A and v0B = 0. So, the equation gets reduced to this:
1/2 · aA · t² = x0B + 1/2 · aB · t²
If we replace with the data we have and solve for t:
1/2 · 4 m/s² · t² = 2900 m - 1/2 · 11 m/s² · t²
2 m/s² · t² = 2900 m - 5.5 m/s² · t²
5.5 m/s² · t² + 2 m/s² · t² = 2900 m
7.5 m/s² · t² = 2900 m
t² = 2900 m / 7.5 m/s²
t = 19.7 s
Car B reaches car A in 19.7 s.
Car B reaches car A in 19.67 seconds.
How do you calculate the time?
Given that the acceleration of car A is 4 m/s2 and car B is -11 m/s2. Let us consider that the initial position of car A is 0 m and the initial position of car B is 2900 m.
The position of an object which is moving in a straight line at constant speed is given by the formula,
[tex]x = x_0 + v_0 t + \dfrac {1}{2}at^2[/tex]
Where x is the position of the object at time t, x_0 is the initial position, v_0 is the initial velocity and a is the acceleration of the object.
The cars will meet at one meeting point, their position will be the same at the time of the meeting.
[tex]x_a = x_b[/tex]
[tex]x_{a0} + v_{a0} t + \dfrac {1}{2}a_at^2 = x_{b0} + v_{b0} t + \dfrac {1}{2}a_bt^2[/tex]
At the meeting point, both the cars are at rest, hence their initial speed v_a0 and v_b0 will be zero.
Hence substituting the values in the above equation.
[tex]0 + 0 + \dfrac {1}{2} \times 4 \times t^2 = 2900 + 0 + \dfrac {1}{2} \times (-11) \times t^2[/tex]
[tex]2t^2 = 2900 -5.5t^2[/tex]
[tex]7.5t^2 = 2900\\[/tex]
[tex]t^2 = 386.67[/tex]
[tex]t = 19.67 \;\rm s[/tex]
Hence we can conclude that car B reaches car A in 19.67 seconds.
To know more about the position of an object, follow the link given below.
https://brainly.com/question/3072589.
