Question

The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 54 km/s . To the crew's great surprise, a Klingon ship is 110 km directly ahead, traveling in the same direction at a mere 22 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 3.4 s. The Enterprise's computers react instantly to brake the ship. What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship?

Answer 1

To avoid a collision, the Starship Enterprise needs to decelerate at a magnitude of -9410 m/s², converting from 54 km/s to 22 km/s in 3.4 seconds.

Step-by-step explanation:

In order to avoid colliding with the Klingon ship, the Starship Enterprise needs to reduce its speed from 54 km/s to 22 km/s to match the speed of the Klingon ship. Given the short amount of time (3.4 seconds) before a potential collision, we can calculate the required acceleration (or deceleration in this case) using the kinematic equation:

v_f = v_i + a ⋅ t

where:

• v_f is the final velocity (22 km/s)
• v_i is the initial velocity (54 km/s)
• a is the acceleration (what we are solving for)
• t is the time (3.4 seconds)

Rearranging the equation to solve for acceleration, we get:

a = (v_f - v_i) / t

Now, we just plug in the values and solve for acceleration:

a = (22 km/s - 54 km/s) / 3.4 s

a = (-32 km/s) / 3.4 s

a = -9.41 km/s²

Since we're discussing acceleration in terms of meters per second squared, we need to convert kilometers per second squared to meters per second squared:

a = -9.41 km/s² × 1000 m/km

a = -9410 m/s²

The negative sign indicates a deceleration, which is what the Enterprise needs to do to avoid the collision.

Answer 2

The acceleration will be "-5.12 km/s²".

Step-by-step explanation:

The given values are:

This might be defined as a continuous inanimate object at either the speed between such organization as well as the Klingon boat (54 - 22) km/s.

The values given in the question are:

= 110 km

Final velocity will be:

= 0

By using to the third equation of motion, we get

⇒
`v^2=u^2+2as`

`0=(54-22)^2+2* a* 100`

`0=(32)^2+2* a* 100`

`0=1024+2* a* 100`

`0=1024+200a`

`-1024=200a`

`a=-(1024)/(200)`

`=-5.12 \ km/s^2`