Answer:
(a) 8.117 (b) 0.742[tex]m/sec^2[/tex]
Explanation:
We have given distance s =40 m time t=7.5 sec
Final velocity v =2.55 m/sec
From the first equation of motion [tex]v=u-at[/tex] (negative sign because there is retrdation as the truck speed is slowing down )
So [tex]2.55=u-7.5a[/tex] --------------eqn 1
From the second equation of motion [tex]s=ut-\frac{1}{2}at^2[/tex] ( negative sign because there is retrdation as the truck speed is slowing down )
So [tex]40=7.5u-0.5\times a\times 7.5^2[/tex]
[tex]40=7.5u-28.125a[/tex]------------------eqn 2
On solving eqn1 and eqn 2
u=8.117 m/sec and a=-0.742[tex]m/sec^2[/tex]
