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given initial velocity 10m/s and final velocity 15m/s calculate the distance traveled in 4 seconds , the distance traveled in last 2sec and the distance traveled in 3rd and 4th seconds​

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Final answer:

The distance traveled in 4 seconds is 50 meters. The distance traveled in the last 2 seconds is 25 meters. The distance traveled in the 3rd and 4th seconds is also 25 meters.

Step-by-step explanation:

To calculate the distance traveled in 4 seconds, we can use the formula:



distance = (initial velocity + final velocity) / 2 * time



Substituting the given values, distance = (10 + 15) / 2 * 4 = 50 meters.



For the distance traveled in the last 2 seconds, we need to find the initial and final velocities in those 2 seconds. Since the acceleration is not provided, we assume it to be constant throughout the entire motion. Using the formula:



final velocity = initial velocity + acceleration * time



We can rearrange the formula to find the acceleration:



acceleration = (final velocity - initial velocity) / time



Substituting the given values, acceleration = (15 - 10) / 4 = 1.25 m/s^2.



We can then find the final velocity in the last 2 seconds:



final velocity = initial velocity + acceleration * time = 10 + 1.25 * 2 = 12.5 m/s.



Finally, we can calculate the distance traveled in the last 2 seconds using the formula:



distance = (initial velocity + final velocity) / 2 * time = (10 + 12.5) / 2 * 2 = 25 meters.



The distance traveled in the 3rd and 4th seconds is simply the difference between the distances traveled in 4 seconds and the last 2 seconds:



distance = 50 - 25 = 25 meters.

User Cskwrd
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