An automobile accelerates from rest at
0.9 m/s
2
for 25 s. The speed is then held
constant for 18 s, after which there is an acceleration
of −5.8 m/s
2 until the automobile
stops.
What total distance was traveled?
Answer in units of km
Well, let me crunch the numbers for you (while being careful not to crunch any gears).
First, let's calculate the distance traveled during the initial acceleration. We can use the equation:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the initial velocity is 0 m/s, the acceleration is 0.9 m/s^2, and the time is 25 seconds, we have:
distance1 = (0 * 25) + (0.5 * 0.9 * 25^2)
Calculating this gives us a distance1 of 281.25 meters.
Next, let's calculate the distance traveled while maintaining a constant speed. Since the speed is constant, there is no acceleration, so the distance is simply:
distance2 = speed * time = 0.9 m/s * 18 s = 16.2 meters.
Lastly, let's calculate the distance traveled during the deceleration phase. Again, we can use the same equation:
distance3 = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the initial velocity is the constant speed of 0.9 m/s, the acceleration is -5.8 m/s^2, and the time is 25 seconds, we have:
distance3 = (0.9 * 25) + (0.5 * -5.8 * 25^2)
Calculating this gives us a distance3 of -2287.5 meters.
Since distance can't be negative, we can discard distance3.
The total distance traveled is then:
total distance = distance1 + distance2 = 281.25 meters + 16.2 meters
Converting this to kilometers gives us a total distance of approximately 0.297 kilometers (rounded to three decimal places).
So, the automobile traveled a total distance of around 0.297 kilometers. I hope it didn't run out of jokes on such a short journey!
To find the total distance traveled by the automobile, we need to calculate the distances traveled during the different stages of motion and then sum them up.
First, let's calculate the distance traveled during the acceleration phase. The formula to calculate distance during uniform acceleration is given by:
distance = (initial velocity * time) + (1/2 * acceleration * time^2)
For the first acceleration phase:
Initial velocity (u) = 0 m/s (since the automobile starts from rest)
Acceleration (a) = 0.9 m/s^2
Time (t) = 25 s
distance = (0 * 25) + (1/2 * 0.9 * 25^2)
distance = 0 + (1/2 * 0.9 * 625) = 0 + 281.25 = 281.25 meters
Next, let's calculate the distance traveled during the constant speed phase. Since speed is held constant, the distance traveled can be found by multiplying the speed by time:
Speed = 0.9 m/s (since it was maintained from the previous phase)
Time (t) = 18 s
distance = speed * time = 0.9 * 18 = 16.2 meters
Finally, let's calculate the distance traveled during the deceleration phase. The formula to calculate distance during uniform deceleration (negative acceleration) is the same as that for acceleration:
For the deceleration phase:
Initial velocity (u) = 0.9 m/s (since the speed was held constant)
Acceleration (a) = -5.8 m/s^2 (negative due to the deceleration)
Time (t) = unknown (we'll calculate it)
We know that the automobile comes to a stop during this phase, so the final velocity (v) is 0 m/s. Using the equation of motion v = u + at and rearranging, we can find the time:
0 = 0.9 + (-5.8) * t
5.8t = 0.9
t = 0.9 / 5.8 = 0.155 seconds
Now, we can calculate the distance:
distance = (0.9 * 0.155) + (1/2 * -5.8 * 0.155^2)
distance = 0.1395 + (-0.1127) = 0.0268 meters
Summing up the distances traveled in each phase:
Total distance traveled = distance during acceleration + distance during constant speed + distance during deceleration
Total distance traveled = 281.25 + 16.2 + 0.0268 = 297.4768 meters
Finally, let's convert the distance from meters to kilometers, remembering that 1 kilometer is equal to 1000 meters:
Total distance traveled = 297.4768 meters / 1000 = 0.2974768 kilometers
Therefore, the total distance traveled by the automobile is approximately 0.297 kilometers.