Question

A particle moves along $X-$axis in such a way that its coordinate $\left(X\right)$ varies with time $\left(T\right)$ according to the expression, $x=2-5t+6t^2$. Then the initial velocity of the particle is

Answer 1

Step. 1 Given Data,
$x=2-5t+6t^2$

Step 2. Formula Used,

The displacement is given as,

$x=ut+\frac{1}{2}a{t}^2$, $u$= initial velocity

$a$ = acceleration

$t=$ time

Step 3. Calculating the initial velocity,

Comparing with the given expression,

$$\begin{gathered} x=2-5t+6t^2 \\ x=-5t+6t^2+2 \end{gathered}$$

$x=ut+\frac{1}{2}a{t}^2$

$-5t+6t^2+2=ut+\frac{1}{2}a{t}^2$

Comparing coefficient of t, we get $u=-5m/s$,

$$\begin{gathered} \frac{1}{2}a{t}^2=6t^2 \\ a=12m/s^2 \end{gathered}$$

The initial velocity of the particle is $-5m/s$.