A car, in overtaking trailer, uniformly accelerated its speed from 40 kph to 90 kph in 8 seconds. Calculate the distance traveled during the period of acceleration.
a = 50kph/8s = 25/4 km/hr/s = 1.736 m/s^2
Vo = 40 km/hr = 11.11 m/s, so
s = Vo*t + a/2 t^2
now just plug in the values for Vo,a,t
Well, it seems like the car really wanted to show off its speed to the trailer! Let's calculate the distance traveled during its speedy acceleration.
To calculate the distance traveled during acceleration, we can use the formula:
d = (1/2) * (v1 + v2) * t
where:
d = distance traveled
v1 = initial velocity
v2 = final velocity
t = time taken
Given that the initial velocity (v1) is 40 kph, the final velocity (v2) is 90 kph, and the time taken (t) is 8 seconds, let's calculate the distance.
Plugging in the values into the formula, we have:
d = (1/2) * (40 + 90) * 8
Let's calculate that using my trusty calculator:
d = (1/2) * 130 * 8
d = 0.5 * 130 * 8
d = 65 * 8
d = 520 kph
So, during the period of acceleration, the car traveled a distance of 520 kilometers per hour. That's a pretty impressive distance for a little speeding session!
To calculate the distance traveled during the period of acceleration, we can use the equation:
Distance = (Initial Velocity + Final Velocity) / 2 * Time
Given:
Initial velocity (u) = 40 kph
Final velocity (v) = 90 kph
Time (t) = 8 seconds
First, let's convert the velocities from kph to m/s:
Initial velocity (u) = 40 kph * (1000 m/3600 s) = 11.11 m/s (approximately)
Final velocity (v) = 90 kph * (1000 m/3600 s) = 25 m/s (approximately)
Substituting the values in the equation:
Distance = (11.11 m/s + 25 m/s) / 2 * 8 seconds
Distance = (36.11 m/s) / 2 * 8 seconds
Distance = 288.88 meters (approximately)
Therefore, the distance traveled during the period of acceleration is approximately 288.88 meters.
To calculate the distance traveled during the period of acceleration, we can use the formula:
\[Distance = \frac{1}{2} \times \text{acceleration} \times \text{time}^2\]
First, we need to calculate the acceleration. The acceleration can be determined by using the formula:
\[Acceleration = \frac{\text{change in velocity}}{\text{time}}\]
Given that the initial velocity (\(u\)) is 40 kph, the final velocity (\(v\)) is 90 kph, and the time (\(t\)) is 8 seconds, we can substitute these values into the formula and calculate the acceleration.
\[Acceleration = \frac{v - u}{t}\]
\[Acceleration = \frac{90 - 40}{8}\]
Calculating the acceleration:
\[Acceleration = \frac{50}{8} = 6.25 \text{ kph/s}\]
Now that we have the acceleration, we can substitute it along with the time into the formula for distance:
\[Distance = \frac{1}{2} \times \text{acceleration} \times \text{time}^2\]
\[Distance = \frac{1}{2} \times 6.25 \times (8)^2\]
Calculating the distance:
\[Distance = \frac{1}{2} \times 6.25 \times 64 = 200 \text{ meters}\]
Therefore, the distance traveled during the period of acceleration is 200 meters.