Q20P - Fundamentals Of Physics

Question

(a) If the position of a particle is given by $\mathbf{x}\mathbf{=}\mathbf{20}\mathbf{t}\mathbf{-}\mathbf{5}{\mathbf{t}}^{\mathbf{2}}$, where x is in m and t in sec when, if ever, the particle’s velocity is zero? (b) When is its acceleration a zero? (c) For what time range (positive or negative) is a negative? (d) Positive? (e) Graph x(t), v(t) and a(t).

Step-by-Step Solution

Answer

(a) Particle’s velocity is zero at $t=1.2 s$

(b) Particle’s acceleration is zero at $t=0 s$

(c) Range is negative at $t>0$

(d) Range is positive at $t<0$

(e) Graph of $x\left(t\right)$,$v\left(t\right)$ and $a\left(t\right)$

Step 1: Given information

$x\left(t\right)=20t-5t^3$

Step 2: To understand the concept of simple mathematical operation

This problem involves simple mathematical operation of derivative. Here differentiating the given equation, the equation for velocity and acceleration can be found. Further, by plugging  and, time where  and  are zero can be calculated.

 

Formula:

$v\left(t\right)=\frac{dx\left(t\right)}{dt}$

Step 3: Derive the expressions for velocity and acceleration

$x\left(t\right)=20t-5t^2$

Expressions for velocity and acceleration can be found by taking derivative of above equation with respect to t,

$$\begin{gathered} v\left(t\right)=\frac{dx\left(t\right)}{dt} \\ =\frac{d\left(20t-5t^2\right)}{dt} \\ =20-15t^2 \\ a\left(t\right)=\frac{dv\left(t\right)}{dt} \\ =\frac{d\left(20-15t^2\right)}{dt} \\ =-30t \end{gathered}$$

Step 4: (a) Calculate the time at which velocity is zero

$$\begin{gathered} v\left(t\right)=20-15t^2 \\ 0=20-15t^2 \end{gathered}$$

$$\begin{gathered} t=\sqrt{\frac{20}{15}} \\ =1.2 s \end{gathered}$$

Step 5: (b) Calculate the time at which acceleration is zero

$$\begin{gathered} a\left(t\right)=-30t \\ 0=-30t \\ t=0s \end{gathered}$$

Step 6: (c)Calculate the time range in which acceleration is negative

From equation, if acceleration $\mathrm{a}\left(\mathrm{t}\right)= -30\mathrm{t}$at $t>0$ acceleration is negative.

Step 7: (d) Calculate the time range in which acceleration is positive

From equation, if acceleration$a\left(t\right)= -30t$ at $t<0$ acceleration is positive.

Step 8: (e) Graph x(t), v(t) and a(t)

Graphs are shown below: