Video Transcript
If a particle was projected from
the origin 𝑂 on a horizontal ground with initial velocity 17𝐢 plus 11𝐣 metres per
second, write its position vector at time 𝑡 seconds. Consider the acceleration due to
gravity to be 9.8 metres per square second.
We’re given the initial velocity of
the particle in a vector form. Its horizontal component is 17
metres per second and its vertical component is 11 metres per second. We want to find its position vector
— let’s call that 𝑠 — at time 𝑡 seconds. And so we’re going to use one of
the equations of constant acceleration. We’re going to use the equation 𝑠
equals 𝑢𝑡 plus a half 𝑎𝑡 squared. Now, we can convert this into
vector form. We can say that 𝑠, 𝑢, and 𝑎 are
vectors.
Lets substitute what we know about
our particle into this formula. 𝑠 is equal to 𝑢 times 𝑡. And we know that 𝑢 is 17𝐢 plus
11𝐣. And then we have a half 𝑎𝑡
squared. Now, acceleration in the horizontal
direction is zero. But in the vertical direction, it
acts against the direction of the original velocity. So it’s negative 9.8. And so we find that 𝑠 is equal to
17𝐢 plus 11𝐣𝑡 plus a half times zero 𝐢 minus 9.8𝐣𝑡 squared.
Let’s distribute each of these set
of parentheses. And we get 𝑠 equals 17𝑡𝐢 plus
11𝑡𝐣 minus 4.9𝐣𝑡 squared. And we now gather the vertical
components. And that gives us 𝑠 equals 17𝑡𝐢
plus 11𝑡 minus 4.9𝑡 squared 𝐣. This is measured in metres. And so we’re done. We found a position vector at time
𝑡 seconds. Its 𝑠 equals 17𝑡𝐢 plus 11𝑡
minus 4.9𝑡 squared 𝐣 metres.
