Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s.

Answer 1

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  $X = 38.3 \ m$

Explanation:

From the question we are told that

The acceleration along the x axis is  $a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)$

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  $v_{0}x = 7.10 m/s$

Generally from the equation for acceleration along x axis we have that

     $a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)$

=>   $\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt$

=>   $V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

At  t =0  s   and  $v_{0}x = 7.10 m/s$

=>   $7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1$

=>   $K_1 = 7.10$      

So  

      $\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1$

=>  $\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}$

=>  $X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2$

At  t =0  s   and   x = -14.0 m

  $-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2$

=>   $K_2 = -14$

So

     $X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14$

At  t = 10.0 s

      $X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14$

=>   $X = 38.3 \ m$