To find the time taken for an object to travel a certain distance with a constant acceleration, we can use the following equation of motion:
s = ut + (1/2)at²
Where: s is the distance covered (20 m), u is the initial velocity (10 m/s), a is the acceleration (5 m/s²), t is the time taken (unknown).
We need to rearrange the equation to solve for t:
t² + (2u/a)t - (2s/a) = 0
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
For our equation, the coefficients are: a = 1, b = 2u/a = 2(10 m/s)/(5 m/s²) = 4 s, c = -2s/a = -2(20 m)/(5 m/s²) = -8 s².
Substituting these values into the quadratic formula:
t = (-4 ± √(4² - 4(1)(-8))) / (2(1)) t = (-4 ± √(16 + 32)) / 2 t = (-4 ± √48) / 2 t = (-4 ± 4√3) / 2
Simplifying further:
t = -2 ± 2√3
Since time cannot be negative in this context, we can ignore the negative solution.
Therefore, the time taken for the object to travel 20 m with a constant acceleration of 5 m/s² is:
t = -2 + 2√3
Approximately:
t ≈ 0.464 s