A bus travelling of 15m/s.accelerates uniformly at 4m/s2.what is the distance covered in log
To find the distance covered in this scenario, we can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Given that the initial velocity (u) is 15 m/s and the acceleration (a) is 4 m/s^2, we can substitute these values into the equation:
distance = 15 * t + (1/2) * 4 * t^2
We want to find the distance covered when the bus is travelling at logarithmic speed, which means that at some point the speed of the bus will become log(10) = 1.
To find the time it takes for the bus to reach a logarithmic speed, we can rearrange the equation:
1 = 15 * t + (1/2) * 4 * t^2
Rearranging this equation gives us a quadratic equation:
2t^2 + 15t - 1 = 0
We can solve this quadratic equation to find the value of t. Once we know t, we can substitute it back into the distance formula to find the distance covered.
Using the quadratic formula:
t = (-15 ± √(15^2 - 4*2*(-1))) / (2*2)
Simplifying this equation gives us:
t = (-15 ± √(225 + 8)) / 4
t = (-15 ± √(233)) / 4
We take the positive root because we are looking for a positive time value.
t ≈ 0.25 seconds
Now, we can substitute this value back into the distance formula:
distance = 15 * 0.25 + (1/2) * 4 * (0.25)^2
Simplifying this equation gives us:
distance ≈ 3.75 + (1/2) * 4 * 0.0625
distance ≈ 3.75 + 0.125
distance ≈ 3.875 meters
Therefore, the distance covered by the bus when it is traveling at a logarithmic speed is approximately 3.875 meters.