Question

A space craft flying in a straight course with a velocity of 75 km/s fires its rocket motors for 6 s. At the end of this time, its speed is 120 km/s in the same direction. Find: 1) the space craft's average acceleration while the motors were firing. 2) 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 s.

Answer 1

Answer:Here , u = 75000 m/s or 7.5×10^(4) m/s

& v = 120 × 1000 m/s or 12×10^(4) m/s

t = 6 s . Let average accerleration be a

Now, a = (v-u)/t =((12-7.5)×10^(4))÷6 = 0.75 ×10^4 m/s^(2) or 7.5 km/s^(2).

b) For this part the distance is calculated in two parts ,d = d1+ d2

Here, d1= distance during first 6 s.

=> d1 = ut+ at^(2)/2

=> d1 = 585 km ( putting u= 7.5 × 10^(4) , t = 6 s & a = 0.75×10^4 )

Similarly d2 = 480km ( putting u = 12 × 10^(4) , a = -10 & t= 4s )

So, d = 585 +480 =1065 km.