Question

A space shuttle traveling at a speed of 95m/s begins to accelerate negative using its parachute and braking system and comes to rest in 500m. Determine the time required to bring the shuttle to rest.​

Answer 1

Answer:

Given:

Initial velocity of space shuttle = 95 m/s

Distance travelled after which it stopped = 500 m

To find:

Time required to bring the shuttle to rest

Concept:

First we will try to find out the acceleration (retardation) of the space shuttle using equations of Kinematics. Then using the retardation value, we will find out the time taken.

Calculation:

Let final velocity be v , initial velocity be u , acceleration be a , time be t and distance travelled be s

${v}^{2} = {u}^{2} + 2as \\ = > {0}^{2} = {(95)}^{2} + 2a(500) \\ = > 1000a = - 9025 \\ = > a = - 9.025 \: m {s}^{ - 2}$

Now we know that :

$v = u + at \\ = > 0 = 95 + ( - 9.025)t \\ = > t = \frac{95}{9.025 } \\ = > t = 10.526 \: sec$

So final answer is :

$\boxed{ \red{ \huge{ \bold{time = 10.526 \: s}}}}$