The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.0 m/s . Initially, the car is also traveling at a speed 19.0 m/s and its front bumper is a distance 24.3 m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.640 m/s2 , then pulls back into the truck's lane when the rear of the car is a distance 25.7 m ahead of the front of the truck. The car is of length 4.40 m and the truck is of length 21.7 m .How much time is required for the car to pass the truck?

15.8 sec

To determine the time required for the car to pass the truck, we need to find the time it takes for the car to reach the position where its rear is 25.7 m ahead of the front of the truck.

First, let's find the initial separation between the car and the truck. The front bumper of the car is initially 24.3 m behind the truck's rear bumper.

Next, let's find the distance the car needs to cover in order to pass the truck. The length of the car is 4.40 m, and the length of the truck is 21.7 m. So, the total distance the car needs to cover is the sum of the lengths of the car and the truck: 4.40 m + 21.7 m = 26.10 m.

Now, let's determine the acceleration of the car. The car starts with an initial speed of 19.0 m/s and accelerates at a constant rate of 0.640 m/s^2.

We can use the kinematic equation to find the time it takes for the car to cover the required distance. The equation is:

d = v_i * t + 1/2 * a * t^2

Where:
- d is the distance covered
- v_i is the initial velocity
- a is the constant acceleration
- t is the time

In this case, we want to find the time (t) when the displacement (d) is 26.10 m, the initial velocity (v_i) is 19.0 m/s, and the acceleration (a) is 0.640 m/s^2.

Rearranging the equation, we have:

t = (-v_i +/- sqrt(v_i^2 + 2 * a * d)) / a

Now, we can substitute the values into the equation and solve for t:

t = (-19.0 +/- sqrt(19.0^2 + 2 * 0.640 * 26.10)) / 0.640

After calculating, we find two solutions for t: t1 and t2.

t1 ≈ 23.63 s (where we choose the negative square root in the equation)
t2 ≈ 0.482 s (where we choose the positive square root in the equation)

Since t represents time and time cannot be negative in this context, we discard t1. Therefore, the car takes approximately t2 ≈ 0.482 s to pass the truck.