Question

If \[x = {t^2} - 5t + 2\] for one particle, Find the velocity at $t = 5s$ ?

Answer 1

Hint: To solve this type of problem we should know about the velocity, instantaneous velocity, and most important is how to find instantaneous velocity. So, to find it we should know about the basic differentiation.
Velocity: the rate of change of displacement with respect to time is called velocity. It is a vector quantity and its S.I. unit is $m{s^{ - 1}}$ .
Distance: Change is position of object from fixed point with time. It’s S.I. unit is $m$ . Distance will always be positive.
Instantaneous velocity: The velocity at a particular point of time is called instantaneous velocity.
 Formula used:
 $v$ $ = $ $\dfrac{d}{t}$
 $y = \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$

Complete step by step solution:
As we have to find velocity at $t = 5s$ .
So, as given distance is the function of time \[x = {t^2} - 5t + 2\] .
Differentiation of $x$ with respect to time will be velocity.
As, \[v = \dfrac{{dx}}{{dt}}\]
So, \[v = \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}({t^2} - 5t + 2)\]
By doing differentiation, we get:
 \[v = 2t - 5\]
So, we find the value of velocity in the function of time, so we get \[v = 2t - 5\] .
Keeping the value of $t = 5s$ .
We get,
 \[v = 2 \times 5 - 5 = 5m{s^{ - 1}}\] .

Hence the velocity of the particle at $t = 5s$ will be \[v = 5m{s^{ - 1}}\] .

Note: if we have to find velocity between $t = o\,\,and\,t = 5\operatorname{s} $ Then we will first find position at $t = o$ and then at $t = 5\operatorname{s} $ .So, subtract and distance travel by particle in given time and divide it from given time and we will get the velocity.