Question

The velocity of a particle moving along a straight line is given by v(t)=1.3tln(0.2t+0.4) for time t≥0. What is the acceleration of the particle at time t=1.2 ?

Answer 1

To find the acceleration of the particle at time t=1.2, differentiate the velocity function v(t)=1.3tln(0.2t+0.4) with respect to time. Plug in t=1.2 into the acceleration function a(t) = 1.3(0.2t+0.4)(1/t)+1.3ln(0.2t+0.4) to calculate the acceleration.

Step-by-step explanation:

To find the acceleration of the particle at time t=1.2, we need to differentiate the velocity function v(t)=1.3tln(0.2t+0.4) with respect to time. Differentiating the function gives us the acceleration function a(t)=1.3(0.2t+0.4)(1/t)+1.3ln(0.2t+0.4). Plugging in t=1.2 into the acceleration function gives us the acceleration of the particle at time t=1.2.

Let's calculate the acceleration:

a(t) = 1.3(0.2t+0.4)(1/t)+1.3ln(0.2t+0.4)

a(1.2) = 1.3(0.2(1.2)+0.4)(1/1.2)+1.3ln(0.2(1.2)+0.4)

a(1.2) = 1.3(0.24+0.4)(1/1.2)+1.3ln(0.24+0.4)

a(1.2) = 1.3(0.64)(1/1.2)+1.3ln(0.64)

a(1.2) = 0.83+1.3ln(0.64)

Therefore, the acceleration of the particle at time t=1.2 is approximately 0.83+1.3ln(0.64).

Answer 2

The acceleration of the particle at time t=1.2 is approximately 2.12.

Find the first derivative:

`v^(\prime)(t)=1.3\left((0.2 t)/((1)/(5)(t+2))+\ln (0.2 t+0.4)\right)`

Find the second derivative:

`v^(\prime \prime)(t)=(1.3(t+t \ln (0.2 t+0.4)+2 \ln (0.2 t+0.4)))/(t+2)`

Evaluate the second derivative at
`$\mathrm{t}=1.2$` :

`\begin{aligned}& v^(\prime \prime)(1.2)=(1.3(1.2+1.2 \ln (0.2 \cdot 1.2+0.4)+2 \ln (0.2 \cdot 1.2+0.4)))/(1.2+2) \\& v^(\prime \prime)(1.2) \approx 2.12\end{aligned}`