Question

Part AA space vehicle accelerates uniformly from 80m/s att 0 to 167 m/s att 10.0 sHow far did it move between t = 2.0 s and t - 6.08 ?Express your answer to two significant figures and include the appropriate units.

Answer 1

Given:

Velocity at t = 0: 80 m/s

Velocity at t = 10.0 s: 167 m/s

Let's find the distance it covered between t = 2.0s and t = 6.0 s

Apply the kinematics formula:

`v=u+at`

Where:

v is the final velocity ==> 167 m/s

u is the initial velocity ==> 80 m/s

a is the acceleration

t is the time ==> 10.0 - 0 = 10.0 s

Now, let's find the acceleration.

Rewrite the formula for a:

`\begin{gathered} a=(v-u)/(t) \\ \\ a=(167-80)/(10) \\ \\ a=(87)/(10) \\ \\ a=8.7m/s^2 \end{gathered}`

The acceleration is 8.7 m/s²

To find the distance, apply the formula:

`s=ut+(1)/(2)at^2`

Thus, we have:

• At t = 2.0 s:

`\begin{gathered} d_1=80(2)+(1)/(2)(8.7)(2)^2 \\ \\ d_1=160+17.4 \\ \\ d_1=177.4\text{ m} \end{gathered}`

• At t = 6.0 s:

`\begin{gathered} d_2=ut+(1)/(2)at^2 \\ \\ d_2=80(6)+(1)/(2)(8.7)(6)^2 \\ \\ d_2=480+156.6 \\ \\ d_2=636.6\text{ m} \end{gathered}`

To find the distance traveled between t = 2.0 s and t = 6.0 s, we have:

d = d2 - d1 = 636.6 m - 177.4 m

d = 459.2 m

Therefore, the distance traveled betwen t = 2.0 s and t = 6.0 s is 459.2 m

ANSWER:

459.2 m