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A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car's motion, the eastward component of the car's velocity is given by vx(t)=(0.890m/s3)t2.
Part A
What is the acceleration of the car whenvx = 14.1 m/s ?
Express your answer with the appropriate units.

ax=

Respuesta :

Answer:

  a_x(3.98) = 7.0844 m/s^2  

Explanation:

Given:

- The eastward component of car's velocity is given as follows:

                                         v_x (t) = 0.890*t^2

For,                                    0 s < t < 5 s

Find:

What is the acceleration of the car when v_x = 14.1 m/s ?

Solution:

- We will first compute the time t at which the velocity v_x component equals 14.1 m /s :

                                         14.1 = 0.890*t^2

                                           t = sqrt (15.8427)

                                           t = 3.98 s

- Now use the relation of v_x and derivative to obtain a_x acceleration in eastward direction:

                                         a_x (t) = dv_x / dt

                                         a_x(t) = 1.78*t

- Evaluate at t = 3.98 s:

                                          a_x(3.98) = 1.783.98

                                          a_x(3.98) = 7.0844 m/s^2  

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