1) The acceleration is [tex]0.96 m/s^2[/tex]
2) The time taken is 39.6 s
Explanation:
1)
Since the motion of the truck is a uniformly accelerated motion (=constant acceleration), we can use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the truck in this problem,
u = 0 (it starts from rest)
v = 38 m/s
s = 755 m
Solving for a, we find the acceleration:
[tex]a=\frac{v^2-u^2}{2s}=\frac{38^2-0}{2(755)}=0.96 m/s^2[/tex]
2)
For this part we can use the following suvat equation
[tex]v=u+at[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken for the velocity to change from u to v
In this problem,
u = 0
v = 38 m/s
[tex]a=0.96 m/s^2[/tex]
Solving for t,
[tex]t=\frac{v-u}{a}=\frac{38-0}{0.96}=39.6 s[/tex]
Learn more about accelerated motion:
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