Question

1) A plane lands on a runway with a speed of 125 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m/s2) and direction of the plane's average acceleration during this time interval?2) Influenced by the gravitational pull of a distant star, the velocity of an asteroid changes from from +20.8 km/s to −17.1 km/s over a period of 2.01 years.(a) What is the total change in the asteroid's velocity? (Indicate the direction with the sign of your answer.)(b) What is the asteroid's average acceleration during this interval? (Indicate the direction with the sign of your answer.)

Answer 1

Answer: 1) a = 9.61m/s² pointing to west.

             2) (a) Δv = - 37.9km/s

                  (b) a = - 6.10⁷km/years

Explanation: Aceleration is the change in velocity over change in time.

1) For the plane:

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

$a=\frac{125}{13}$

a = 9.61m/s²

The plane is moving east, so velocity points in that direction. However, it is stopping at the time of 13s, so acceleration's direction is in the opposite direction. Therefore, acceleration points towards west.

2) Total change of velocity:

$\Delta v = v_{f}-v_{i}$

$\Delta v = -17.1-(+20.8)$

$\Delta v= -37.9$km/s

The interval is in years, so transforming seconds in years:

v = $\frac{-37.9}{3.15.10^{-7}}$

$v=-12.03.10^{7}$km/years

Calculating acceleration:

$a=\frac{-12.03.10^{-7}}{2.01}$

$a=-6.10^{7}$

Acceleration of an asteroid is a = -6.10⁷km/years .