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
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
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!
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.
\[Distance = \frac{1}{2} \times \text{acceleration} \times \text{time}^2\]
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.