A car is behind a truck going 23.0 m/s on the highway. The driver looks for an opportunity to pass, guessing that his car can accelerate at 2.2 m/s2, and he gauges that he has to cover the 24.0 m length of the truck, plus 10 m clear room at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably traveling at the speed limit of 25 m/s. He estimates that the car is about 900.0 m away.
a) How long does it take the driver to complete the pass?
b) If x1 is the position of the driver, and x2 is the position of the oncoming car, what would be the value of separation distance (x2 - x1) at the end of a successful pass?
c) What would be the driver's speed at the end of the pass?
I am not sure what equations to use?
So, he has to go 44 m further than the truck in the same time. He has to do this before the car meets them.
900m= 25t + 23t +1/2 2.2 t^2
solve for t. check that.
I solved for the quadratic and calculated 14.157 seconds. This assignment is on-line, so I get instant feedback and was told that this answer was incorrect. Any explanation? Thank you so much for the help.
To solve this problem, we can use the concept of relative motion. Let's break down the problem into several steps:
Step 1: Calculate the time required for the driver to pass the truck.
The driver needs to cover a distance of 24.0 m (length of the truck), plus 10 m (clear room at the rear of the truck), and another 10 m (clear room at the front of the truck). So, the total distance the driver needs to cover is (24.0 + 10 + 10) = 44.0 m more than the distance covered by the truck.
Since the driver is guessing that his car can accelerate at 2.2 m/s^2, we can use the equation of motion:
distance = initial velocity * time + 0.5 * acceleration * time^2
Let's consider the motion of the car relative to the truck. The truck's velocity is 23.0 m/s, and the car's velocity relative to the truck is given by:
relative velocity = car's velocity - truck's velocity
relative velocity = 0 m/s (since both are moving at the same velocity)
Now, let's calculate the time required for the car to cover the additional distance:
44.0 = 0 * t + 0.5 * 2.2 * t^2
Simplifying the equation gives:
1.1 * t^2 = 44.0
t^2 = 40.0
t ≈ 6.324 s
Therefore, it takes approximately 6.32 seconds for the driver to complete the pass.
Step 2: Calculate the separation distance (x2 - x1) at the end of a successful pass.
To calculate the separation distance, we need to consider the distance traveled by both the oncoming car and the driver's car during the time it takes for the pass.
Given that the oncoming car is approximately 900.0 m away and is traveling at a speed of 25 m/s, we can use the equation of motion for the oncoming car:
distance = initial velocity * time
900 = 25 * t
Solving for t gives:
t = 900 / 25
t = 36.0 s
Now, let's calculate the separation distance (x2 - x1) at the end of the pass:
(x2 - x1) = (driver's velocity relative to the truck) * t
The driver's velocity relative to the truck can be calculated using the equation:
velocity = initial velocity + acceleration * time
velocity = 0 + 2.2 * 6.324
velocity ≈ 13.92 m/s
(x2 - x1) = 13.92 * 36.0
(x2 - x1) ≈ 501.12 m
Therefore, the separation distance (x2 - x1) at the end of a successful pass is approximately 501.12 meters.
Step 3: Calculate the driver's speed at the end of the pass.
To calculate the driver's speed at the end of the pass, we need to add the driver's initial velocity to the relative velocity between the car and the truck.
The driver's initial velocity is the same as the truck's velocity, which is 23.0 m/s. The relative velocity between the car and the truck is 0 m/s since both are moving at the same velocity.
Therefore, the driver's speed at the end of the pass is 23.0 m/s.
In summary:
a) The driver takes approximately 6.32 seconds to complete the pass.
b) The separation distance (x2 - x1) at the end of a successful pass is approximately 501.12 meters.
c) The driver's speed at the end of the pass is 23.0 m/s.