<h3>

</h3><h3>Given</h3>
v = 20m\s
a = 3m\s^2
t = 4sec
Firstly we have to find u
a = 
3m\s =
12m\s = 20 - u
20 - u = 12m\s
- u = -8
u = 8
Now we can easily find distance by using second equation of motion
s = ut + 1\2 at^2
s = 8(4) + 1\2(3)(16)
s = 32 + 24
s = 56
So distance is 56 m\s hope it helps
Answer:
Acceleration, 
Explanation:
Initial velocity of a particle in vector form, u = (-5i - 2j) m/s
Final velocity of particle in vector form, v = (-6i + 7j) m/s
Time taken, t = 8 seconds
We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :

or

Hence, the value of acceleration vector is solved.
C Weight is the gravitational pull on an object
Speed v = distance travelled / time taken
v = d / t
v = 540 / 60h
v = 9 km /h
The energy absorbed by photon is 1.24 eV.
This is the perfect answer.