Question

A spaceship of proper length ? = 100 m travels in the positive x direction at a speed of 0.700 c0 relative to Earth. An identical spaceship travels in the negative x direction along a parallel course at the same speed relative to Earth. At t = 0, an observer on Earth measures a distance d = 58,000 km separating the two ships.

Part A

At what instant does this observer see the leading edges of the two ships pass each other?

Answer 1

The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.

Step-by-step explanation:

Given that,

Length = 100 m

Speed = 0.700 c

Distance = 58000 km

The distance should be halved because the spaceships both travel the same speed.

So they will meet at the middle of the distance

We need to calculate the distance

Using formula for distance

`d'=(d)/(2)`

Put the value into the formula

`d'=(58000)/(2)`

`d'=29000\ km`

We need to calculate the time at which the observer see the leading edges of the two ships pass each other

Using formula of time

`\Delta t=(d)/(V)`

Put the value into the formula

`\Delta t=(29*10^(6))/(0.700*3*10^(8))`

`\Delta t=0.138\ sec`

Hence, The time at which the observer see the leading edges of the two ships pass each other is 0.138 sec.

Answer 2

observer see the leading edges of the two ships pass each other at time 0.136 s

Step-by-step explanation:

given data

spaceship length = 100 m

speed of 0.700 Co =
`0.700* 3 * 10^8` m/s

distance d = 58,000 km = 58000 × 10³ m

solution

as here distance will be half because both spaceship travel with same velocity

so they meet at half of distance

Distance Da =
`(d)/(2)` ............1

Distance Da =
`(58000* 10^3)/(2)`

Distance Da = 29 ×
`10^(6)` m

and

time at which observer see leading edge of 2 spaceship pass

Δ time =
`(Da)/(v)` ..........2

Δ time =
`(29 * 10^6)/(0.700* 3 * 10^8)`

Δ time = 0.136 s