An increasing net force is applied to a 20 kg spacecraft in intergalactic space.

The spacecraft is traveling at 25.0 m/s in the direction of the applied force after 2.0 s. It's velocity after 6.0 s is a.bc x 10^{d }m/s. The values for a, b, c, and d are _, _, _, and _.

Attached Graph:

9 months ago

Answered By Jackie C

Solution:

The acceleration of the spacecraft is A=(v1-v0)/t1=25.0 m/s/2s=12.5 m/s^2. Assuming its acceleration is constant, we have

A=(v2-v0)/t2, v2=A*t2+vo=12.5 m/s^2*6.0 s+o.0=12.5*6.0+0=75.0 m/s = 7.50*10^1 m/s, then

a=7

b=5

c=0, and

d=1

9 months ago

Answered By Denis B

Since v= 25 m/s after 2.0 sec., we can calculate acceleration by the formula a=v/t; therefore, a=25 m/s / 2.0 sec. which gives us a = 12.5 m/sec^2. Question: what will v be after 6.0 secs.?

Solution - v is given by the formula v=at (acceleration multiplied by time) where a is 12.5 m/sec.^2 and time is 6.0 secs. Therfore v= [12.5 x 6] m/s or 75 m/s and, in scientific notation is 7.50 x 10^1 m/s

a=7, b=5, c=0, and d=1 !! Voila! Physics is the best... enjoy!!

9 months ago

## Answered By Jackie C

Solution:

The acceleration of the spacecraft is A=(v1-v0)/t1=25.0 m/s/2s=12.5 m/s^2. Assuming its acceleration is constant, we have

A=(v2-v0)/t2, v2=A*t2+vo=12.5 m/s^2*6.0 s+o.0=12.5*6.0+0=75.0 m/s = 7.50*10^1 m/s, then

a=7

b=5

c=0, and

d=1

9 months ago

## Answered By Denis B

Since v= 25 m/s after 2.0 sec., we can calculate acceleration by the formula a=v/t; therefore, a=25 m/s / 2.0 sec. which gives us a = 12.5 m/sec^2. Question: what will v be after 6.0 secs.?

Solution - v is given by the formula v=at (acceleration multiplied by time) where a is 12.5 m/sec.^2 and time is 6.0 secs. Therfore v= [12.5 x 6] m/s or 75 m/s and, in scientific notation is 7.50 x 10^1 m/s

a=7, b=5, c=0, and d=1 !! Voila! Physics is the best... enjoy!!