two cars start from rest and begin accerlating, one at 10m/s^2 and the other at 12m/s^2. how long will it take for the second car to be 100m ahead of the first car
just solve for t in
5t^2 + 100 = 6t^2
To determine the time it takes for the second car to be 100m ahead of the first car, we can use the equations of linear motion.
Let's assume that the first car has an initial position (displacement) of 0m, an initial velocity of 0m/s, and an acceleration of 10m/s².
For the first car:
Initial displacement (s₁) = 0m
Initial velocity (v₁) = 0m/s
Acceleration (a₁) = 10m/s²
The equations of motion for the first car are:
s₁ = v₁t + (1/2)a₁t² -- Equation 1
v₁ = a₁t -- Equation 2
Similarly, let's assume that the second car also starts from rest but has an acceleration of 12m/s².
For the second car:
Initial displacement (s₂) = 0m
Initial velocity (v₂) = 0m/s
Acceleration (a₂) = 12m/s²
The equations of motion for the second car are:
s₂ = v₂t + (1/2)a₂t² -- Equation 3
v₂ = a₂t -- Equation 4
We want to find the time (t) it takes for the second car to be 100m ahead of the first car. This means the displacement of the second car (s₂) will be 100m greater than the displacement of the first car (s₁).
Therefore, s₂ - s₁ = 100m
Substituting the values into the equations, we have:
(1/2)a₂t² - (1/2)a₁t² = 100m
Simplifying the equation further:
(1/2)(a₂ - a₁)t² = 100m
(1/2)(12m/s² - 10m/s²)t² = 100m
t² = (100m) / (1/2)(12m/s² - 10m/s²)
t² = (100m) / (1/2)(2m/s²)
t² = (100m) / (2m/s²)
t² = 50s²
t = sqrt(50s²)
t ≈ 7.071s
Therefore, it will take approximately 7.071 seconds for the second car to be 100m ahead of the first car.