A man starts with a velocity of 15m/s and moves with an acceleration of 2m/s^2.what distance does he cover in the fifth second? What distance does he cover in 10s
his distance, after t seconds, is
s(t) = 15t + t^2
The 5th second is the time between the end of the 4th second and the end of the 5th second. So, the distance covered is
s(5)-s(4) = (15*5+5^2) - (15*4+4^2) = 100-76 = 24m
The distance covered in the first 10 seconds is thus
s(10)-s(0) = s(10) = 250m
To find the distance a man covers in a specific time, we need to use the equations of motion in physics. In this case, we can use the equation:
distance = initial velocity * time + (0.5 * acceleration * time^2)
Let's calculate the distance covered in the fifth second:
Given:
Initial velocity (u) = 15 m/s
Acceleration (a) = 2 m/s^2
Time (t) = 5 seconds
Plugging these values into the equation, we have:
distance = (15 * 5) + (0.5 * 2 * (5^2))
distance = 75 + (0.5 * 2 * 25)
distance = 75 + 25
distance = 100 meters
Therefore, the man covers a distance of 100 meters in the fifth second.
Now, let's calculate the distance covered in 10 seconds:
Given:
Initial velocity (u) = 15 m/s
Acceleration (a) = 2 m/s^2
Time (t) = 10 seconds
Using the same equation:
distance = (15 * 10) + (0.5 * 2 * (10^2))
distance = 150 + (0.5 * 2 * 100)
distance = 150 + 100
distance = 250 meters
Therefore, the man covers a distance of 250 meters in 10 seconds.