During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, $3 \times {10^8}m{s^{ - 1}}$.
Answer
592.5k+ views
Hint – Here we will proceed by using the formula for speed to find out the distance of the spaceship from the ground station.
Formula used - $Speed = \dfrac{{Dis\tan cetravelled}}{{Timetaken}}$
Complete step-by-step answer:
Here it is given that,
$3 \times {10^8}m{s^{ - 1}}$
Given, the signal travels at the speed of light,
$v = 3 \times {10^8}m{s^{ - 1}}$
We know that,
Time taken by the signal to reach the ground $ = 5\min $
$ = 5 \times 60\sec $
$ = 300\sec $
Let the distance of the spaceship from the ground station be $D m$
We know,
$Speed = \dfrac{{Dis\tan cetravelled}}{{Timetaken}}$
$
\Rightarrow v = \dfrac{d}{t} \\
\Rightarrow D = v \times t \\
$
$
= 3 \times {10^8} \times 300 \\
= 900 \times {10^8} \\
= 9 \times 100 \times {10^8} \\
= 9 \times {10^{8 + 2}} \\
= 9 \times {10^{10}}m \\
$
So the distance of the spaceship from the ground is $9 \times {10^{10}}$ meters.
Note – Whenever we come up with this type of question, where we are asked to find out whether the distance or speed. Then we first write the formula, after writing the formula we will put the values. After solving that we will find out the asked quantity (here distance of a spaceship from a ground station).
Formula used - $Speed = \dfrac{{Dis\tan cetravelled}}{{Timetaken}}$
Complete step-by-step answer:
Here it is given that,
$3 \times {10^8}m{s^{ - 1}}$
Given, the signal travels at the speed of light,
$v = 3 \times {10^8}m{s^{ - 1}}$
We know that,
Time taken by the signal to reach the ground $ = 5\min $
$ = 5 \times 60\sec $
$ = 300\sec $
Let the distance of the spaceship from the ground station be $D m$
We know,
$Speed = \dfrac{{Dis\tan cetravelled}}{{Timetaken}}$
$
\Rightarrow v = \dfrac{d}{t} \\
\Rightarrow D = v \times t \\
$
$
= 3 \times {10^8} \times 300 \\
= 900 \times {10^8} \\
= 9 \times 100 \times {10^8} \\
= 9 \times {10^{8 + 2}} \\
= 9 \times {10^{10}}m \\
$
So the distance of the spaceship from the ground is $9 \times {10^{10}}$ meters.
Note – Whenever we come up with this type of question, where we are asked to find out whether the distance or speed. Then we first write the formula, after writing the formula we will put the values. After solving that we will find out the asked quantity (here distance of a spaceship from a ground station).
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

