Question

v2=
a. 4²-2as b.4²+2as c. 4+2as d. 4²+as​

Answer 1

• Answer:
• We will use both of the equations of motion to reach the third equation of motion. This will require a bit of algebra.
• S=ut+
• 2
• 1
• at
• 2
• andv=u+at, include the time variant t
• There will be some situations when we do not have any information about time and so it would be a good idea to derive an equation that does not have a t term.
• To do this, we rearrange our first equation to get
• t=
• a
• v−u
• and use this to replace t wherever it appears in the second equation. So
• S=ut+
• 2
• 1
• at
• 2
• becomes,
• S=u(
• a
• v−u
• )+
• 2
• 1
• a(
• a
• v−u
• )
• 2
• ⇒2aS=2u(v−u)+(v−u)
• 2
• ⇒2aS=2uv−2u
• 2
• −v
• 2
• −2uv−u
• 2
• ⇒2aS=v
• 2
• −u
• 2
• ⇒v
• 2
• =u
• 2
• +2aS

Answer 2

Answer:

a. 4²+2as

Explanation:

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