. A car traveling at 25 mi/h accelerates to 48 mi/h over the first 5 seconds. It maintains that speed for the next 5 seconds, and then slows to a stop during the next 5 seconds
What would it look as a grpgh?
final speed = 0
easy.
just apply
s(t) = s0</sub + v0</subt + 1/2 at^2
s(t) = s0 + v0t + 1/2 at^2
paha maia
answeer pl0x
To answer this question, we need to calculate the acceleration of the car. Acceleration is the rate at which the velocity of an object changes. In this case, the car goes from 25 mi/h to 48 mi/h over a period of 5 seconds.
To calculate the acceleration, we can use the formula:
Acceleration (a) = (Final Velocity (Vf) - Initial Velocity (Vi)) / Time (t)
Given:
Vi = 25 mi/h
Vf = 48 mi/h
t = 5 seconds
Substituting the values into the formula, we get:
Acceleration (a) = (48 mi/h - 25 mi/h) / 5 s
Acceleration (a) = 23 mi/h / 5 s
So, the acceleration of the car is 4.6 mi/h/s.
Now, let's find out the distance traveled during each of the given time intervals.
First 5 seconds:
To calculate the distance traveled during the first 5 seconds, we need to use the formula:
Distance (d) = Initial Velocity (Vi) * Time (t) + 0.5 * Acceleration (a) * Time (t)^2
Given:
Vi = 25 mi/h
t = 5 seconds
a = 4.6 mi/h/s
Substituting the values into the formula, we get:
Distance (d) = 25 mi/h * 5 s + 0.5 * 4.6 mi/h/s * (5 s)^2
Distance (d) = 125 mi + 0.5 * 4.6 mi/h/s * 25 s^2
Distance (d) = 125 mi + 0.5 * 4.6 * 625 mi
Distance (d) = 125 mi + 1437.5 mi
Distance (d) = 1562.5 mi
The car travels a distance of 1562.5 mi during the first 5 seconds.
Next 5 seconds:
During the next 5 seconds, the car maintains a constant speed of 48 mi/h. Therefore, the distance traveled during this interval can be calculated simply by multiplying the velocity by the time:
Distance (d) = Velocity (V) * Time (t)
Distance (d) = 48 mi/h * 5 s
Distance (d) = 240 mi
The car travels a distance of 240 mi during the next 5 seconds.
Last 5 seconds:
To find the distance the car travels during the last 5 seconds, we need to calculate the deceleration. The car is coming to a stop, so its final velocity will be 0 mi/h.
Using the formula for acceleration, we can find the deceleration:
Acceleration (a) = (Final Velocity (Vf) - Initial Velocity (Vi)) / Time (t)
0 mi/h = (0 mi/h - 48 mi/h) / 5 s
Simplifying the equation, we have:
-48 mi/h = -48 mi/h / 5 s
The negative sign indicates that the car is decelerating. Therefore, the deceleration is 48 mi/h divided by 5 s, which is equal to 9.6 mi/h/s.
Now, we can find the distance traveled during the last 5 seconds using the formula:
Distance (d) = Initial Velocity (Vi) * Time (t) + 0.5 * Deceleration (a) * Time (t)^2
Given:
Vi = 48 mi/h (since the car maintains that speed during the previous 5 seconds)
t = 5 seconds
a = -9.6 mi/h/s
Substituting the values into the formula, we get:
Distance (d) = 48 mi/h * 5 s + 0.5 * -9.6 mi/h/s * (5 s)^2
Distance (d) = 240 mi + 0.5 * -9.6 mi/h/s * 25 s^2
Distance (d) = 240 mi + 0.5 * -9.6 * 625 mi
Distance (d) = 240 mi - 2880 mi
Distance (d) = -2640 mi
The negative sign indicates that the car is moving backward. Therefore, the car travels a distance of 2640 mi backward during the last 5 seconds.
To summarize:
- The car travels 1562.5 mi during the first 5 seconds.
- The car travels 240 mi during the next 5 seconds.
- The car travels 2640 mi backward during the last 5 seconds.
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