Respuesta :
Explanation:
Given that,
Distance, s = 47 m
Time taken, t = 8.6 s
Final speed of the truck, v = 2.3 m/s
Let u is the initial speed of the truck and a is its acceleration such that :
[tex]a=\dfrac{v-u}{t}[/tex].............(1)
Now, the second equation of motion is :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Put the value of a in above equation as :
[tex]s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2[/tex]
[tex]s=\dfrac{t(u+v)}{2}[/tex]
[tex]u=\dfrac{2s}{t}-v[/tex]
[tex]u=\dfrac{2\times 47}{8.6}-2.3[/tex]
u = 8.63 m/s
So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.
