Data:
• Position equation
[tex]x(t)=1.9t^2[/tex]
The exercise is asking for speed.
The first derivative of position is speed.
Procedure:
1. Calculating the first derivative: obtaining the equation of speed v ( t )
[tex]\begin{gathered} v(t)=\frac{dx}{dt} \\ v(t)=1.9\cdot2t \\ v(t)=3.8t \end{gathered}[/tex]
2. Calculating the time it takes to reach 45.6m/s: isolating t
[tex]\begin{gathered} v(t)=3.8t \\ t=\frac{v(t)}{3.8} \\ t=\frac{45.6}{3.8} \\ t=12 \end{gathered}[/tex]
Answer: 12s
Summary:
0. Calculating the first derivative of position, which equals to speed
,
1. Isolating ,t
,
2. Obtaining the time
