Answer:
(A) Original velocity will be 7.244 m /sec
(B) Acceleration will be [tex]-0.5026m/sec^2[/tex]
Explanation:
We have given distance covers by truck s = 40 m
Time taken by truck to cover this distance t = 7.45 sec
Final velocity v = 3.50 sec
According to second equation of motion
[tex]S=ut+\frac{1}{2}at^2[/tex]
[tex]40=u\times 7.45+\frac{1}{2}\times a\times 7.45^2[/tex]
[tex]7.45u+27.751a=40[/tex]-----eqn 1
According to first equation of motion
v = u + at
So [tex]3.5=u+7.45a[/tex]-----eqn2
Solving equation 1 and 2
a = [tex]-0.5026m/sec^2[/tex]
And u = 7.244 m /sec
(A) Original velocity will be 7.244 m /sec
(B) Acceleration will be
[tex]-0.5026m/sec^2[/tex]
