Question

what is the resultant velocity of a point of x-component Vx = t^3 - 1 and y-component Vy = t^2 - t at time t =4?

Answer 1

With the use of formula, the resultant velocity is 64 m/s approximately.

Velocity

Velocity is the distance travelled in a specific direction per time taken. Velocity is a vector quantity. It can be represented in a vector diagram.

The S.I unit of velocity is meter per second. That is, m/s

Given that the coordinate velocity  of a point of x-component Vx = t^3 - 1 and y-component Vy = t^2 - t at time t = 4. The resultant velocity of this point can be calculated by using the below formula.

$V_{R}$ = $\sqrt{V_{x} ^{2} + V_{y} ^{2} }$

Where

• $V_{R}$ = Resultant Velocity

• $V_{x}$ = x - component of the velocity

• $V_{y}$ = y - component of the velocity

Since time t = 4s.

For x - component,

$V_{x}$ = $t^{3}$ - 1

$V_{x}$ = $4^{3}$ - 1

$V_{x}$ = 64 - 1

$V_{x}$ = 63 m/s

For y - component

$V_{y}$ = $t^{2}$ - t

$V_{y}$ = $4^{2}$ - 4

$V_{y}$ = 16 - 4

$V_{y}$ = 12 m/s

$V_{R}$ = $\sqrt{V_{x} ^{2} + V_{y} ^{2} }$

$V_{R}$ = $\sqrt{63^{2} + 12^{2} }$

$V_{R}$ = $\sqrt{4113}$

$V_{R}$ = 64.13 m/s

Therefore, the resultant velocity on that point is 64.13 m/s

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