Question

the displacement of x of a particle moving in One dimension under the action of constant force is related to the time by the equation X =(5t ^2 + 3t-5 ) m where x is in metres and T is in seconds. find the velocity of the particle at 1(t) = 3 s 2(t)=6s ​

Answer 1

Answer:

Please find the attachment

formula used as follow of differentiation

$\frac{d}{dx} {x}^{n} = n {x}^{n - 1} \\ \frac{d}{dx} x = 1 \\ \frac{d}{dx} k = 0 \\ and \: velocity \: = \frac{d}{dt} x \: where \: x \: is \: distance \: covered$

Answer 2

Answer:

Please find the attachment

formula used as follow of differentiation

\begin{gathered} \frac{d}{dx} {x}^{n} = n {x}^{n - 1} \\ \frac{d}{dx} x = 1 \\ \frac{d}{dx} k = 0 \\ and \: velocity \: = \frac{d}{dt} x \: where \: x \: is \: distance \: covered\end{gathered}

dx

d

x

n

=nx

n−1

dx

d

x=1

dx

d

k=0

andvelocity=

dt

d

xwherexisdistancecovered