When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters from its starting position in 8.3 s.
Why did the car start braking? Did it see a sign that said "Stop and smell the roses"? Or did it realize it left the oven on at home? Either way, we should applaud its initiative for stopping before it ended up in a neighboring state. Now, let's crunch some numbers, shall we?
The change in position (Δx) is given by the final position minus the initial position. In this case, Δx = 350 - 100 = 250 meters. The time (Δt) it took to come to a stop is 8.3 seconds. Using the formula Δx = v initial * Δt + (1/2) * a * Δt^2, where v initial is the initial velocity and a is the acceleration, we can solve for a. Rearranging the formula, we get: a = (Δx - v initial * Δt) / (1/2 * Δt^2). Plugging in the given values, we have a = (250 - 60 * 8.3) / (1/2 * 8.3^2). Now, let's calculate that acceleration! *taps calculator* Voila! The car had an acceleration of approximately -14.277 m/s^2. Negative because it's slowing down!
To solve this problem, let's break it down into steps:
Step 1: Calculate the initial velocity (u) of the car.
Given:
Distance (s) = 100 meters
Time (t) = 8.3 seconds
Final velocity (v) = 0 m/s
We can use the formula: v = u + at, where a is the acceleration.
Rearranging the formula, we have: u = (v - at)
Substituting the values, we get:
u = (0 - (0 * 8.3))
u = 0 m/s
Therefore, the initial velocity (u) of the car is 0 m/s.
Step 2: Calculate the acceleration (a) of the car.
Since the car starts braking to come to a stop, the final velocity (v) is 0 m/s.
We can use the formula: v^2 = u^2 + 2as
Rearranging the formula, we have: a = (v^2 - u^2) / (2s)
Substituting the values, we get:
a = (0^2 - 60^2) / (2 * 350)
a = (-3600) / 700
a ≈ -5.14 m/s^2
NOTE: The negative sign indicates that the acceleration is in the opposite direction of the initial motion.
Therefore, the acceleration (a) of the car is approximately -5.14 m/s^2.
Step 3: Calculate the deceleration (decel) of the car.
Since the acceleration is negative due to braking, the deceleration is the absolute value of the acceleration:
decel = |a|
decel = |-5.14|
decel ≈ 5.14 m/s^2
Therefore, the deceleration (decel) of the car is approximately 5.14 m/s^2.
Overall, the initial velocity (u) of the car is 0 m/s, the acceleration (a) is approximately -5.14 m/s^2, and the deceleration (decel) is approximately 5.14 m/s^2.
To solve this problem, we will use the equations of motion. We are given the initial position, final position, initial velocity, and time. We need to find the acceleration.
Let's start by listing the given values:
- Initial position (x0) = 0 meters
- Final position (xf) = 350 meters
- Initial velocity (v0) = 60.0 m/s
- Time (t) = 8.3 seconds
We can use the equation of motion to find the acceleration (a) and the final velocity (vf). The equation is:
xf = x0 + v0t + (1/2)at^2
Rearranging the equation to solve for acceleration, we get:
a = (2(xf - x0) - 2v0t) / t^2
Plugging in the values, we have:
a = (2(350 - 0) - 2(60.0)(8.3)) / (8.3)^2
Simplifying this equation will give us the value of acceleration.
a = (700 - 996) / 68.89
a = -296 / 68.89
a ≈ -4.293 m/s^2
Therefore, the car has a deceleration of approximately 4.293 m/s^2.