Question

A space craft flying in a straight course with a velocity of 75 $m{s}^{-1}$ fires its rocket motors for 6.0 s.At the end of this time, its speed is 120 $km{s}^{-1}$ in the same direction.Find:(i) the space craft's average acceleration while the motors were firing,(ii) the distance travelled by the space craft in the first 10 s after the rocket motors were started,the motors having been in action for only 6.0 s.

Answer 1

Use kinematic equations:
s-displacement, u-initial velocity, v-final velocity,a- acceleration, t-time

v=u + at
$v^2=u^2+2as$
s=$\frac{1}{2}$(u+v)t
s=ut +$\frac{1}{2}a{t}^2$

a)Calculate the average acceleration when the motors were firing:
change in velocity = 120 - 75 = 45 m/s
av. acceleration= $\frac{changeinvelocity}{time}$ = $\frac{45}{6}$= 7.5 $m/s^2$
therefore the average acceleration = 7.5 $km/s^2$

b)Distance traveled after the rockets were fired:
Find distance traveled during the 6s when the motors were firing:
t=6s, v=120 km/s, u=75 km/s, a=7.5 $km/s^2$

s= $\frac{1}{2}$(u+v)t
= $\frac{1}{2}$(120+75)6
=$\frac{1}{2}$ x 195 x 6
= 585 km
Find distance after the motors are fired:(the next 4 seconds to make it 10 seconds)
After the motors are fired, the craft moves with 120km/s speed
That means in 4s = 4 x 120
= 480km
add the distance traveled in the 10s = (585 + 480)km
= 1065km
Therefore the total distance traveled = 1065 km