Question

A spacecraft flying in a straight course at 75 kms-1 fires its rocket motors for 6.0 seconds. At the end of 6.0 seconds its speed becomes 120 kms-1 in the same direction. The spacecraft’s average acceleration while the motors were firing is​

Answer 1

Answer:

acceleration is 7.5m/s^2.

Explanation:

Answer 2

Answer:

Use kinematic equations:

SUVAT

s-displacement, u-initial velocity, v-final velocity,a- acceleration, t-time

v=u + at

v2 = u2 + 2as

s=1/2(u+v)t

s=ut + 1/2at2

a)Calculate the average acceleration when the motors were firing:

av. acceleration = change in velocity   =    120 - 75    =  45m/s   = 7.5m/s2

                                         time                              6s              6s

therefore the average acceleration = 7.5km/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=75km/s, a=7.5km/s2

 s= 1/2(u+v)t

 = 1/2(120+75)6

 =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