Question

Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +19.9 km/s to −18.3 km/s over a period of 1.97 years.
(a) What is the total change in the asteroid's velocity? (Indicate the direction with the sign of your answer.) m/s
(b) What is the asteroid's average acceleration during this interval? (Indicate the direction with the sign of your answer.)

Answer 1

Answer:

Given:

Initial velocity of the asteroid, v = 19.9 km/s = 19900 m/s

(since, 1km = 1000 m)

Final velocity of the asteroid, v' = - 18.3 km/s = -18300 m/s

time period, T = 1.97 years

Solution:

(a). Total change in velocity, $\Delat v$:

$\Delat v$ = v' - v

$\Delat v$ = - 18300 - 19900 = - 38200 m

here, negative sign is indicative of the fact that the motion of the asteroid is in the downward direction.

(b) Asteroid's average acceleration, a:

a = $\frac{\Delta v}{T}$

a = $\frac{- 38200}{1.97\times 365\times 24\times 3600}$

a = $- 6.15\times 10^{- 4} m/s^{2}$

Here, the negative sign is the indication of deceleration of the asteroid in the downward direction.