Question

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Answer 1

★ Given :

initial velocity = 75 km/s

final velocity = 120 km/s

rocket motors are fired for - 6 s

★ To find :

i) Average acceleration

ii) distance travelled by spacecraft in first 10 s after rocket motors are started.

★ Solution :

i) v = u + at

by putting values

120 = 75 + 6a

6a = 120 - 75

6a = 45

$a = \frac{45}{6} \\ \\ \\ = \bold{7.5\: km/s^2}$

ii) divide the distance travelled in two parts :

$d_1 \: for \: first \: 6 \: seconds \\ \\ \\ d_2 \: for \: last\: 4 \: seconds \\ \\ \\ \bold{ s = ut + \frac{1}{2}at^2} \\ \\ \\ d_1 = (75 \times 6) + \frac{1}{2} \times 7.5 \times 6 \times 6 \\ \\ \\ = 450 + 135 \\ \\ \\ = 585 \: km$

$d_2 = v \times t \\ \\ \\ = 120 \times 4 \\ \\ \\ = 480 \\ \\ \\ d = d_1 + d_2 \\ \\ \\ = 585 + 480 = 1065 \: km \\ \\ \\ \therefore \: total\: distance = 1065 \: km$

* Refer to the attachment for 2nd part.

Answer 2

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A space craft flying in a straight course with a velocity of 75 km s·¹ . At fires it's rocket motors for 6-0 s. At the end of this time, it's speed is 120 km s·¹ in the same direction. Find : (i) the space crafts average acceleration while the motors were firing, (ii) the distance travelled by the space craft in the first 10s after the rocket motors were started, the motors having been in action for only 6·0 s ?

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★ Given :

initial velocity = 75 km/s

final velocity = 120 km/s

Rocket motors are fired for - 6 s

★ To find :

i) Average acceleration

ii) distance travelled by spacecraft in first 10 s after rocket motors are started.

★ Solution :

i) v = u + at

by putting values

120 = 75 + 6a

6a = 120 - 756a = 45

$\begin{gathered} a = \frac{45}{6} \\ \\ \\ = \bold{7.5\: km/s^2} \end{gathered}$

a= 645

=>7.5km/s 2

ii) divide the distance travelled in two parts :

$\begin{gathered} d_1 \: for \: first \: 6 \: seconds \\ \\ \\ d_2 \: for \: last\: 4 \: seconds \\ \\ \\ \bold{ s = ut + \frac{1}{2}at^2} \\ \\ \\ d_1 = (75 \times 6) + \frac{1}{2} \times 7.5 \times 6 \times 6 \\ \\ \\ = 450 + 135 \\ \\ \\ = 585 \: km \end{gathered} </p><p>d 1 forfirst6secondsd 2</p><p> forlast4seconds</p><p>s=>ut+ 21at 2d 1 =(75×6)+ 21 ×7.5×6×6</p><p>=>450+135</p><p>=>585km</p><p> \begin{gathered} d_2 = v \times t \\ \\ \\ = 120 \times 4 \\ \\ \\ = 480 \\ \\ \\ d = d_1 + d_2 \\ \\ \\ = 585 + 480 = 1065 \: km \\ \\ \\ \therefore \: total\: distance = 1065 \: km\end{gathered} </p><p>d 2 =v×t</p><p>=>120×4</p><p>=>480d=d 1 +d 2$

==>585+480=1065km

∴totaldistance=1065km

* Refer to the attachment for 2nd part.