Question

Question #c843a

Answer 1

`v=25ms^-1`

Explanation:

`m=1500kg`
`v_(0)=15ms^-1`
`F_1=1000N`
`t=15s`

Before a force acting on it, the object is at constant velocity (assuming no other forces is acting on it: `eg.` friction), thus
`NetF=0`

After a force acting on it, the object experiences a net force of same magnitude as the applied force.
`F=ma`
`1000=1500a`
`a=2/3 ms^-2`

`v=v_0+at`
`v=15+2/3*15`
`v=15+10`
`v=25ms^-1`

Answer 2

Its velocity will be 25m/s

Explanation:

We can use two known equations to solve this problem.
These are
`F=m*a`
`v=u +a*t`

Lets substitute the values
F=1000N
m=1500 kg
then we have `a=F/m=1000/1500` =2/3 m/`s^2`

Also
u=15 m/s
t=15 s
so that `v=u+a*t=15 + 2/3*15=15+10=25s`