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,

$120 - 75 = a \times 6 \\[0.5em] 45 = 6a \\[0.5em] \Rightarrow a = \dfrac{45}{6} \\[0.5em] \Rightarrow a = 7.5 \text { km s}^{-2} \\[0.5em]$

Hence, acceleration = 7.5 km s^-2

(ii) As we know,

S = ut + $\dfrac{1}{2}$at²

For the first 6 s,

u = 75 km s^-1

a = 7.5 km s^-2

$S${1} = (75 \times 6) + (\dfrac{1}{2} \times 7.5 \times 6^2) \\[0.5em] S{1} = 450 + (\dfrac{1}{2} \times 7.5 \times 36) \\[0.5em] S{1} = 450 + (7.5 \times 18) \\[0.5em] S{1} = 450 + 135 \\[0.5em] \Rightarrow S_{1} = 585 \text { 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,

$S$2 = 120 \times 4 \\[0.5em] S2 = 480 \text { 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.