Video Transcript
A car is traveling east and passes the point P that is 45 meters east of a road junction. As the car passes the point P, its speed is 30 meters per second and the driver applies the brakes, accelerating west at 2.5 meters per second squared. What distance east of the road junction is the car 10 seconds after the brakes are applied?
To work through this question, it’ll be very helpful to draw a diagram. We have the junction, the point labeled P, and a set of axes to help show the cardinal directions. Here, we’ve drawn the car as it is passing point P. Remember that the car is heading to the east. But the moment it passes point P, the car begins to brake, or accelerate toward the west. We’ll label this moment as 𝑡 equals zero seconds. In other words, this is the beginning moment at which we will model the car’s motion.
We were told that the car accelerates at a constant rate of 2.5 meters per second squared. So we can use the following equation of motion to describe its movement. 𝑠 equals 𝑣 times 𝑡 plus one-half 𝑎 times 𝑡 squared, where 𝑠 is the car’s displacement. 𝑣 is the car’s initial velocity as it passes point P. 𝑎 is the car’s acceleration. And 𝑡 is the time over which the car accelerates.
We should keep in mind that because we’re using the moment that the car passes point 𝑃 as the start time, this equation will give us the car’s displacement 𝑠 as measured from point P. But to reach our final answer, we need to find the car’s displacement from the road junction.
It’s also important to note that displacement, velocity, and acceleration are all vector quantities. And therefore, we need to define a positive direction and a negative direction before we start to substitute values into this equation. Since the car is traveling east and we want to calculate the total distance traveled east of the road junction, it’ll make sense for us to choose the eastward direction to be positive.
Now we know that as it passes point P, the car’s initial velocity 𝑣 has a value of 30 meters per second to the east. So we can say that 𝑣 equals positive 30 meters per second. When the brakes are applied, the car accelerates at 2.5 meters per second squared to the west. Since we defined the eastward direction to be positive, the acceleration of the car is negative, as the acceleration points to the west. So acceleration 𝑎 is equal to negative 2.5 meters per second squared.
Now, we want to find the car’s displacement 10 seconds after the brakes are applied. So we’ll use a time value of 10 seconds. We can now solve for the car’s displacement from point P by substituting these values into the equation of motion. Clearing some space on screen, we find that the car’s displacement from point P is equal to 30 meters per second times 10 seconds plus one-half times negative 2.5 meters per second squared times 10 seconds squared, which comes out to 175 meters. This is the car’s displacement east of point P.
But the question is asking us for the total distance from the road junction. We were told that the distance between the road junction and point P is 45 meters. Therefore, the car’s total distance east of the road junction is 175 meters plus 45 meters, which is equal to 220 meters. And so this is our final answer.
The car’s total distance east of the road junction 10 seconds after it passed point P is 220 meters.
