a)
Consider 'a' as the acceleration.
Given that the distance travelled, s = 22.4 m
Initial velocity, u = 53.7 km/h = 14.92 m/s
Time taken, t = 2.84 s
Using the formula, s = ut + 1/2 at2,
22.4 = 14.92 x 2.84 + 0.5 x a x 2.842
4.03a = -19.96
a = - 19.96/4.03
= - 4.95 m/s2
Deceleration = -a = 4.95 m/s2
b)
Using the formula, v = u + at,
Where v is the final velocity.
v = 14.92 + (-4.95) x 2.84
= 0.86 m/s
Chapter 02, Problem A car traveling 53.7 km/h is 22.4 m from a barrier when the...
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