Question
A particle moves along a straight line so that its velocity, \(v{\text{ m}}{{\text{s}}^{ – 1}}\)at time t seconds is given by \(v = 6{{\rm{e}}^{3t}} + 4\) . When \(t = 0\) , the displacement, s, of the particle is 7 metres. Find an expression for s in terms of t.
▶️Answer/Explanation
The velocity of the particle is given by \(v = 6{{\rm{e}}^{3t}} + 4\). We know that velocity is the rate of change of displacement with respect to time, so we can integrate the velocity with respect to time to get the displacement.
$$\begin{aligned} s &= \int_0^t v dt \\ &= \int_0^t (6{{\rm{e}}^{3t}} + 4) dt \\ &= \left[2{{\rm{e}}^{3t}} + 4t\right]_0^t \\ &= 2{{\rm{e}}^{3t}} + 4t – 2 \end{aligned}$$
When \(t = 0\), we have:
$$s = 2{{\rm{e}}^{3 \cdot 0}} + 4 \cdot 0 – 2 = 0$$
However, we know that when \(t = 0\), the displacement of the particle is 7 metres. Therefore, we need to add 7 to the expression we obtained for s:
$$s = 2{{\rm{e}}^{3t}} + 4t – 2 + 7 = 2{{\rm{e}}^{3t}} + 4t + 5$$
Hence, an expression for s in terms of t is \(s = 2{{\rm{e}}^{3t}} + 4t + 5\).
Question
Let \(f'(x)=6x^{2}+2x-1.\) Given that f(2) = 5, find f(x).
Answer/Explanation
Ans:
We have
\(f(x)=\int f'(x)dx\)
\(=\int (6x^{2} + 2x-1)dx\)
\(=2x^{3}+x^{2}-x+C\)
Using the fact that f(2) = 5, we get
\(5=2(2)^{3}+2^{2}-2+C\)
5= 16 + 4 – 2 + C
C = -13
Therefore \(f(x)=2x^{3}+x^{2}-x-13\)