calculusvarying accelerationvelocity and position integration

Straight-line motion with varying acceleration

When the acceleration is not constant but is a known function of time, we can find the velocity and position as functions of time by integrating the acceleration function.

vx=v0x+∫0taxdtxx=x0+∫0tvxdt\begin{aligned} v_x &= v_{0x} + \int_0^t a_x dt\\ x_x &= x_0 + \int_0^t v_x dt \end{aligned}

Exercises

(2.50) A small object moves along the x-axis with acceleration a(t)=−(0.0320m/s3)(15.0s−t)a(t) = -(0.0320m/s^3)(15.0s-t). At t=0t = 0 the object is at x0=−14.0mx_0 = -14.0 m and has velocity v0=4.50m/sv_0 = 4.50 m/s. What is the x-coordinate of the object when t = 10.0 s?

Solution

Identify: Motion with varying acceleration. (Integration)
Setup and Execute:

v(t)=v0+∫a(t)dt=4.5+∫(−0.48+0.032t)dt=0.016t2−0.48t+4.5x(t)=x0+∫0tvdt=−14.0+∫0t(0.016t2−0.48t+4.5)dt=0.0163t3−0.24t2+4.5t−14.0x(10.0)=0.0163(10)3−0.24⋅(10)2+4.5⋅10−14.0=373≈12.3m\begin{aligned} v(t) & = v_0 + \int a(t) dt\\ &= 4.5 + \int (-0.48+0.032t) dt\\ &= 0.016t^2 -0.48t + 4.5\\ x(t) &= x_0 + \int_0^t vdt\\ &= -14.0 + \int_0^t (0.016t^2 -0.48t + 4.5)dt\\ &= \frac{0.016}{3}t^3 -0.24t^2 + 4.5t - 14.0\\ x(10.0) &= \frac{0.016}{3}(10)^3 -0.24 \cdot (10)^2 + 4.5 \cdot 10 - 14.0\\ &= \frac{37}{3} \approx 12.3m \end{aligned}

The acceleration of a particle is given by a(t)=2.00m/s2+(3.00m/s3)ta(t) = 2.00m/s^2 + (3.00m/s^3) t. Find the initial v0v_{0} such that the particle have the x-coordinate, at t=4st=4s and t=0st=0s.

Solution

v(t)=v0+∫0tadt=v0+∫0t(2+3t)dt=v0+2t+32t2x(t)=x0+∫0tvdt=x0+∫0t(v0+2t+32t2)dt=x0+v0t+t2+12t3x(0)=x(t)x0+0=x0+4v0+16+12⋅64⇒v0=−12m/s\begin{aligned} v(t) &= v_0 + \int_0^t adt\\ &= v_0 + \int_0^t (2+3t)dt\\ &= v_0 + 2t + \frac{3}{2}t^2\\ x(t) &= x_0 + \int_0^t vdt\\ &= x_0 + \int_0^t (v_0 + 2t + \frac{3}{2}t^2)dt\\ &= x_0 + v_0 t + t^2 +\frac{1}{2}t^3\\ x(0) &= x(t)\\ x_0 + 0 &= x_0 + 4v_0 + 16 + \frac{1}{2} \cdot 64\\ \To v_0 &= -12m/s \end{aligned}