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We have been given:

final velocity = \(v = 0\text{m/s}\), initial velocity = \(u = 20\text{m/s}\), deceleration = \(a = -2\text{m/s}^2\) (this value is negative because it is decelerating)

\(v = u + at\)

\(0=20+(-2t)\)

\(2t=20\)

\(t=20÷2=10\text{s}\)

v = u + at

where v= final velocity = 0

u = initial velocity = 20m/s

a = retardation = 2m/s²

From the formula:

v = u + at

v - u = at

0 - 20 = 2t

-20 = 2t

Divide both side by co-efficient of T

t = -20/2

t = 10s

Since deceleration is negatively virtual 2

I mean -2.

Given:

\(v = 0\text{m/s}\), \(u = 20\text{m/s}\), \(a = -2\text{m/s}^2\) (the figure is carrying a negative sign because it is decelerating)

\(v = u + at\)

\(0=20+(-2t)\)

\(2t=20\)

\(t=20÷2=10\text{s}\)