Question

Q1.A spaceship travelling in space at 300 km s^{–1}
–1
, fires it's engine for 15 seconds, such that its final velocity is 600 km s^{–1}
–1
. Calculate the total distance travelled by the ship in one minute starting from the time of firing.​

Answer 1

Explanation:

GIVEN :–

• Initial velocity (u) = 300 Km/s

• Final velocity (v) = 600 Km/s

• Time (t) = 15 second

TO FIND :–

• Travelled distance in one minute after fires it's engine ?

SOLUTION :–

• We know that –

v=u+at

• Put the values –

$\begin{gathered} \\ \implies \bf 600= 300+ a(15)\\ \end{gathered} </p><p></p><p> </p><p>$

$\begin{gathered} \\ \implies \bf 600 - 300 = 15a\\ \end{gathered} </p><p>$

$\begin{gathered} \\ \implies \bf 300 = 15a\\ \end{gathered} </p><p>$

$\begin{gathered} \\ \implies \bf a = 20 \: Km/ {s}^{2} \\ \end{gathered} </p><p>$

• So , Travelled Distance in 15 seconds –

$\begin{gathered} \\ \implies \bf S=ut +\dfrac{1}{2} a {t}^{2} \\ \end{gathered} </p><p>$

$\begin{gathered} \\ \implies \bf S_1 =(300)(15)+\dfrac{1}{2} (20) {(15)}^{2} \\ \end{gathered} </p><p>$

$\begin{gathered} \\ \implies \bf S_1 =4500+2250\\ \end{gathered} </p><p>$

$\begin{gathered} \\ \implies \bf S_1 =6750\\ \end{gathered} </p><p>$

• Travelled distance in next 45 sec with uniform speed (no acceleration) is –

$\begin{gathered} \\ \implies \bf S_2 =vt\\ \end{gathered} </p><p>$

$\begin{gathered} \\ \implies \bf S_2 =(600)(45)\\ \end{gathered} </p><p>$

$\begin{gathered} \\ \implies \bf S_2 =27000\\ \end{gathered} </p><p>$

• Hence , Total travelled distance in one minute is –

$\begin{gathered} \\ \implies \bf S = S_1 + S_2\\ \end{gathered} </p><p>$

$\begin{gathered} \\ \implies \bf S = 6750 + 27000\\ \end{gathered} </p><p>$

$\begin{gathered} \\ \implies \large { \boxed{\bf S = 33750 \:\:Km}}\\ \end{gathered} </p><p>$