Answer :
(a) The acceleration of the truck is, [tex]-2.1m/s^2[/tex]
(b) The distance covered by the truck is, 117.8 m
Explanation :
By the 1st equation of motion,
[tex]v=u+at[/tex] ...........(1)
where,
v = final velocity = 20 m/s
u = initial velocity = 30 m/s
t = time = 4.7 s
a = acceleration of the truck = ?
Now put all the given values in the above equation 1, we get:
[tex]20m/s=30m/s+a\times (4.7s)[/tex]
[tex]a=-2.1m/s^2[/tex]
The acceleration of the truck is, [tex]-2.1m/s^2[/tex]
By the 2nd equation of motion,
[tex]s=ut+\frac{1}{2}at^2[/tex] ...........(2)
where,
s = distance covered by the truck = ?
u = initial velocity = 30 m/s
t = time = 4.7 s
a = acceleration of the truck = [tex]-2.1m/s^2[/tex]
Now put all the given values in the above equation 2, we get:
[tex]s=(30m/s)\times (4.7s)+\frac{1}{2}\times (-2.1m/s^2)\times (4.7s)^2[/tex]
By solving the term, we get:
[tex]s=117.8m[/tex]
The distance covered by the truck is, 117.8 m
