The final velocity (v) of the motorcycle is given by the equation:

v = u + at

where:
u is the initial velocity,
a is the acceleration, and
t is the time.

We can rearrange this equation to find the initial velocity:

u = v - at

Substituting the given values:

u = 20 m/s - (1.6 m/s² * t)

We also know that the distance (s) covered by the motorcycle is given by the equation:

s = ut + 0.5at²

Substituting the given values:

50 m = u*t + 0.5*(1.6 m/s²)*t²

We now have two equations with two unknowns (u and t). We can solve these equations simultaneously to find the values of u and t.

First, substitute the expression for u from the first equation into the second equation:

50 m = (20 m/s - 1.6 m/s² * t)*t + 0.5*(1.6 m/s²)*t²

Simplify this equation to find:

50 m = 20 m/s * t - 1.6 m/s² * t² + 0.8 m/s² * t²

Combine like terms:

50 m = 20 m/s * t - 0.8 m/s² * t²

Rearrange this equation to find:

0.8 m/s² * t² - 20 m/s * t + 50 m = 0

This is a quadratic equation in the form of at² - bt + c = 0, where a = 0.8 m/s², b = -20 m/s, and c = 50 m.

We can solve this equation for t using the quadratic formula:

t = [b ± sqrt(b² - 4ac)] / (2a)

t = [20 m/s ± sqrt((-20 m/s)² - 4*0.8 m/s²*50 m)] / (2*0.8 m/s²)

t = [20 m/s ± sqrt(400 m²/s² - 160 m²/s²)] / 1.6 m/s²

t = [20 m/s ± sqrt(240 m²/s²)] / 1.6 m/s²

t = [20 m/s ± 15.49 m/s] / 1.6 m/s²

The two possible solutions for t are:

t = (20 m/s + 15.49 m/s) / 1.6 m/s² = 22.18 s

t = (20 m/s - 15.49 m/s) / 1.6 m/s² = 2.81 s

Since time cannot be negative, the only valid solution is t = 2.81 s.

Substitute t = 2.81 s into the first equation to find the initial velocity:

u = 20 m/s - 1.6 m/s² * 2.81 s = 15.5 m/s

So, the initial velocity of the motorcycle is 15.5 m/s and the time elapsed is 2.81 s.

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