Answer to Question #307672 in Mechanics | Relativity for che

Question

A hovercraft passes point A moving West with a speed of 15 m s−1.

The hovercraft immediately begins to accelerate at a steady rate of 0.13 m s−2 North.

What is the magnitude of the velocity of the hovercraft 200 s after it passes point A? (in m s−1)

Give your answer to 2 s.f.

(note: a hovercraft rides on a cushion of air and so does not contact the ground and experiences no frictional forces)

Answer 1

Explanations & Calculations

• There is acceleration due North and there is no such thing due West.
• So the hovercraft marks a parabolic path just after passing point A.
• This is similar to a horizontal projection under gravity.

• The vertical component of velocity after 200 seconds is

$\qquad\qquad \begin{aligned} \small \uparrow\\ \small v&=\small u+at\\ \small v&=\small 0+0.13\,ms^{-2}\times200\,s\\ &=\small 26\,ms^{-1} \end{aligned}$

• The horizontal component is just the same one as it was at point A.
• Therefore, the magnitude of the velocity is

$\qquad\qquad \begin{aligned} \small |V|&=\small \sqrt{26^2+15^2}=30\,ms^{-1} \end{aligned}$

• Direction of it from West is

$\qquad\qquad \begin{aligned} \small \theta&=\small \tan^{-1}\Big(\frac{26}{15}\Big)=60^0 \,North \,of\,West \end{aligned}$