What force must the brakes apply to a truck of mass 2800 kg moving with a speed of 3 metre per second to bring it to rest in 8 seconds

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Solution

Given,
Initial velocity = 3 m/s
Final velocity = 0 m/s
From the first equation of motion we get,
$v = u + a t$
$u = - a t$
$a = - \frac{u}{t} = - \frac{3}{8}$ (-ve sign shows negative acceleration or retardation)
We know that,
$F o r c e = F = m a$
$F = 2800 \times \frac{- 3}{8}$
$F = - 1050 N$
Negative sign shows that the force applied by the brakes is in opposite direction of the motion of the truck.
Magnitude is 1050 N.

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