An airplane accelerates with a constant rate of 3.0 m/s2 starting at a velocity of 21 m/s. If the final velocity is 60 m/s,

what is the displacement during this period?

526.5m

Explanation:

Given parameters:

Acceleration = 3m/s²

Initial velocity  = 21m/s

Final velocity  = 60m/s

Unknown:

Displacement = ?

Solution:

To solve this problem, we apply the right motion equation which is shown below;

     V²  = U²  + 2aS

  V is the final velocity

  U is the initial velocity

  a is the acceleration

 S is the displacement

 Insert the parameters and solve;

    60²  = 21² + 2(3)S

    3600 = 441 + 6S

    3600 - 441  = 6S

    3159  = 6S

       S = 526.5m

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west