Atruck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) find its original speed.

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 :

.............(1)

Now, the second equation of motion is :

Put the value of a in above equation as :

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//