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

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 =
`(\Delta v)/(T)`

a =
`(- 38200)/(1.97* 365* 24* 3600)`

a =
`- 6.15* 10^(- 4) m/s^(2)`

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