Answer:
Solution below.
Explanation:
First, lets list down all the kinematics equations.
[tex]v = u + at \\ {v}^{2} = {u}^{2} + 2as \\ s = ut + \frac{1}{2} a {t}^{2} [/tex]
where:
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement (or distance(
t is the time taken
We can use any of the equation based on the given variables in the question.
a) Given t = 8 seconds, v = 5 m/s and u = 15 m/s, we can use the first equation to solve for a, the acceleration.
v = u + at
5 = 15 + a × 8
a × 8 + 15 = 5
a × 8 = 5 - 15
a × 8 = - 10
a = -10 ÷ 8
a =
[tex] - 1.25 \frac{m}{ {s}^{2} } [/tex]
(Deccelarating)
b) Now, we can use the acceleration found in part a and the 3rd equation to solve for s, the distance.
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = (15)(8) + \frac{1}{2} ( - 1.25) {(8)}^{2} \\ = 80m[/tex]
