# 41. Let x(t) be the position at time t of a particle moving along the x-axis for t>0. The particle’s position at time t=0 is x=1, and the particle’s velocity at time t is given by x'(t)=t for t>0. at what position on the x-axis is the particle’s when t=4?.

## let x(t) be the position at time t of a particle moving along the x-axis for t>0.

Step-by-step solution-:

The velocity of the particle is given by:

x^{\prime}(t) =t

Let be initial position is xi and the final position be xf:

x_{i}=1, x_{f}=x

Let be initial time is ti and the final time be tf:

t_{i}=0, t_{f}=t

x'(t) is a rate of change of x:

It means,

x^{\prime}(t)=\frac{x_{f}-x_{i}}{t_{f}-t_{i}} :

Substitute the value of ti=0, tf=t into the above expression:

x^{\prime}(t)=\frac{x_{f}-x_{i}}{\textcolor{secondary}{t}-\textcolor{primary}{0}}

Substitute the value of xi=1, xf=x into the above expression:

x^{\prime}(t)=\frac{\textcolor{secondary}{x}-\textcolor{primary}{1}}{{t}-{0}}

Substitute the value of x'(t)=t into the above expression:

\textcolor{primary}{t}=\frac{{x}-{1}}{{t}-{0}}

After cross multiply we get

t\times(t-0)=x-1

After simplifying we get

t^2=x-1

Substitute the t=4 into the above:

4^2=x-1

Therefore, The value of x is:

x=16+1=17

The required position x=17 at t=4 is:

You may be like this post-:

Spread the love