Question

An object moves in a straight line so that its velocity, v m/s , at any time t seconds is given by v(t) = $0.5t^{2} - 2.5t + 2$ .
a)Find the equation that represents the acceleration at any time t.
b)Find the equation representing the position,x metres,at any time t,given that the object is 2.5 metres to the right of the origin when t=0 seconds.

Answer 1

Writing the acceleration and position given the velocity.

Answer:

a) $a(t) = t -2.5$

b) $x(t) = \frac{1}{6}t^3 -\frac{5}{4}t^2 +2t +2.5$

Step-by-step explanation:

Part a:

Acceleration is just the derivative of velocity. In this case, we are given how the velocity, $v(t)$, is defined. If we take the derivative of $v(t)$ that'll be our acceleration.

Recall:

$\frac{\mathrm{d}}{\mathrm{d}x}(x^n) = nx^{n -1}$

$\frac{\mathrm{d}}{\mathrm{d}x}(f(x) +g(x)) = \frac{\mathrm{d}}{\mathrm{d}x}(f(x)) +\frac{\mathrm{d}}{\mathrm{d}x}(g(x))$

$\frac{\mathrm{d}}{\mathrm{d}x}(a \cdot f(x)) = a \cdot \frac{\mathrm{d}}{\mathrm{d}x}(f(x))$

Let $a(t)$ be the acceleration. Solving for $a(t)$:

$v(t) = 0.5t^2 -2.5t +2 \\ \frac{\mathrm{d}}{\mathrm{d}t}(v(t)) = \frac{\mathrm{d}}{\mathrm{d}t}(0.5t^2 -2.5t +2) \\ a(t) = \frac{\mathrm{d}}{\mathrm{d}t}(0.5t^2) -\frac{\mathrm{d}}{\mathrm{d}t}(2.5t) +\frac{\mathrm{d}}{\mathrm{d}t}(2) \\ a(t) = 0.5\frac{\mathrm{d}}{\mathrm{d}t}(t^2) -2.5\frac{\mathrm{d}}{\mathrm{d}t}(t) +\frac{\mathrm{d}}{\mathrm{d}t}(2) \\ a(t) = 0.5(2t) -2.5 +0 \\ a(t) = t -2.5$

Part b:

Position is the same thing as displacement (Although they are fundamentally different but it's okay to treat them as similar things in this case). Displacement is just the integral of velocity. If we take the integral of $v(t)$ that'll be our displacement.

Please refer to my Answer from this question to know the needed information of integrals to solve: brainly.com/question/24687635

Recall:

$\int (a \cdot f(x)) \mathrm{d}x = a\int f(x) \mathrm{d}x$

$\int (f(x) +g(x)) \mathrm{d}x = \int f(x) \mathrm{d}x +\int g(x) \mathrm{d}x$

Let $x(t)$ be our position. Solving for $x(t)$:

$v(t) = 0.5t^2 -2.5t +2 \\ \int v(t) \mathrm{d}t = \int (0.5t^2 -2.5t +2) \mathrm{d}t \\ x(t) = \int (0.5t^2) \mathrm{d}t -\int (2.5t) \mathrm{d}t +\int (2) \mathrm{d}t \\ x(t) = 0.5\int (t^2) \mathrm{d}t -2.5\int (t) \mathrm{d}t +\int (2) \mathrm{d}t \\ x(t) = 0.5(\frac{t^3}{3}) -2.5(\frac{t^2}{2}) +2t +C \\ x(t) = \frac{1}{6}t^3 -\frac{5}{4}t^2 +2t +C$

We still have to solve for $C$. We are given that at $t = 0$, $x(t) = 2.5$. We can therefore write the equation: $x(0) = 2.5$ or $\frac{1}{6}(0)^3 -\frac{5}{4}(0)^2 +2(0) +C = 2.5$.

Solving for $C$:

$\frac{1}{6}(0)^3 -\frac{5}{4}(0)^2 +2(0) +C = 2.5\\ 0 -0 +0 +C = 2.5 \\ C = 2.5$

We can also write the equation:

$x(t) = \frac{1}{6}t^3 -\frac{5}{4}t^2 +2t +2.5$