Answer:
The asteroid's acceleration at this point is [tex]2.71\ m/s^2[/tex]
Explanation:
The equation that governs the trajectory of asteroid is given by :
[tex]x=6.5t-2.3t^3[/tex]
The velocity of asteroid is given by :
[tex]v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2[/tex]
At some point during the trip across the screen, the asteroid is at rest. It means, v = 0
So,
[tex]6.5-6.9t^2=0\\\\t=0.971\ s[/tex]
Acceleration,
[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t[/tex]
Put t = 0.971 s
[tex]a=-13.8\times 0.197\\\\a=-2.71\ m/s^2[/tex]
So, the asteroid's acceleration at this point is [tex]2.71\ m/s^2[/tex] and it is decelerating.
