A space craft flying in a straight course with a velocity of 75 km 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.

(i) As we know,

v – u = at

Given,

u = 75 km s^-1

v = 120 km s^-1

t = 6 s

Substituting the values in the formula above we get,

120 – 75 = a × 6

or, 6a = 45

or, a = 45\over 6

or, a = 7.5 km s^-2

Hence, acceleration = 7.5 km s^-2

(ii) As we know,

S = ut + 1/2 at²

For the first 6 s,

u = 75 km s^-1

a = 7.5 km s^-2

S₁ = (75 × 6) + 1/2 × 7.5 × 6²

= 450 + 1/2 × 7.5 × 36

= 450 + 7.5  × 18

= 450 + 135 = 585 km

Hence, S₁ = 585 km

For the next 4 s

S₂ = speed x time

Given,

t = 4 s

speed = 120 km s ^-1

Substituting the values in the formula above, we get,

S₂ = 120 × 4 = 480 km

Hence, S₂ = 480 km

Total distance covered by the aircraft = S₁ + S₂

= 585 + 480 = 1065 km

Hence,

Total distance covered by the aircraft = 1065 km.