Answer:
No, the deviation in the path of asteroid is not by 22°
Explanation:
Given:
velocity of asteroid, [tex]v_a=21\ km.s^{-1}[/tex]
acceleration of the rocket, [tex]a=0.035\ km.s^{-2}[/tex]
time of acceleration, [tex]t=40\ s[/tex]
Now, the final velocity of the asteroid:
using the equation of motion,
[tex]v=u+a.t[/tex]
where:
[tex]v=[/tex] final velocity
[tex]u=[/tex] initial velocity in the direction
[tex]v=0+0.035\times 40[/tex]
[tex]v=1.4\ km.s^{-1}[/tex]
Now direction of the resultant velocity:
[tex]\tan\beta=\frac{v_a}{v}[/tex]
[tex]\tan\beta=\frac{21}{1.4}[/tex]
[tex]\beta=86.186^{\circ}[/tex]
So, the deviation in the asteroid:
[tex]\theta=90-\beta[/tex]
[tex]\theta=90-86.186[/tex]
[tex]\theta=3.814^{\circ}[/tex]