A rocket traveling at 45.68 m/s is accelerated uniformly to 94.19 m/s over a 10.01 s interval. What is its displacement during this time?

Respuesta :

The displacement of the rocket is 699.5 m

What is the displacement?

Now e know that the term acceleration is the change of velocity with time. We have to obtain the acceleration of the object before we can be able to obtain the distance that have been traveled by the rocket.

Using

v = u + at

v = final velocity

u = initial velocity

a = acceleration

t = time

Then;

94.19 = 45.68 +  10.01a

94.19 - 45.68 =  10.01a

a = 94.19 - 45.68 / 10.01

a = 4.85 m/s^2

Thus;

v^2 = u^2 + 2as

s = v^2 - u^2/2a

s = ( 94.19)^2 - (45.68 )^2/2 * 4.85

s = 6785.1/9.7

s = 699.5 m

Thus, the displacement of the rocket is 699.5 m

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