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.
