A truck is stopped at a stoplight. When the light turns green, it accelerates at 2.2 m/s2. At the same instant, a car passes the truck going 11 m/s. Where and when does the truck catch up with the car?
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well do you know an equation to solve it
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Problem solving is not memorizing equations. It is using equations to find certain elements, then relating those elements to the situation. This is what is required, and there is no substitute.
To find out where and when the truck catches up with the car, we need to analyze their positions and velocities.
Let's denote the initial position of the truck as 'x_t0' and the initial position of the car as 'x_c0'. Also, let's denote the time when the truck catches up with the car as 't' and the final positions of the truck and car as 'x_tf' and 'x_cf', respectively.
Now, let's break down the problem using the following equations of motion:
For the truck:
1. Position of the truck: x_t = x_t0 + v_t0 * t + 0.5 * a_t * t^2
2. Velocity of the truck: v_t = v_t0 + a_t * t
For the car:
3. Position of the car: x_c = x_c0 + v_c * t
Given that the car is already moving at a constant velocity (v_c) when the truck starts to accelerate, we can assume that the car's position remains unchanged during the time interval 't' (since it is not accelerating).
Therefore, we set equations 1 and 3 equal to each other:
x_t0 + v_t0 * t + 0.5 * a_t * t^2 = x_c0 + v_c * t
Now, let's substitute the known values into the equation:
x_t0 + 0 * t + 0.5 * 2.2 * t^2 = x_c0 + 11 * t
Simplifying the equation, we have:
1.1 * t^2 - 11 * t + (x_t0 - x_c0) = 0
This equation is a quadratic equation in terms of 't'. We can solve it to find the time 't' when the truck catches up with the car. Once we know 't', we can substitute it into the equation for the car's position (equation 3) to find the position when the truck catches up with the car.
Please provide the values of x_t0 and x_c0 for further calculations.