I don't know how to solve this problem . Please anyone help. Question:-
An object start moving from immobility with a acceleration of " f m/s^2". At the end of every t seconds it increases it acceleration by "f m/s^2". Show that after "nt seconds" the velocity is "n(n+1)ft/2" and the object has moved (n(n+1)(2n+1)ft^2)/2 meters.
So, Starting at $t=0$ the object has velocity $v(0)=0$ and acceleration $f\ \frac{m}{s^2}$. Thus at time $t$, by the above mentioned formulas, the object will have travelled $$x(t)=\frac12t^2f\ \text{ meters}, $$ and will have a velocity of $$v(t)=ft \ \frac{m}{s}.$$ Between time $t$ and $2t$ it will have travelled, accounting for $\color{red}{\text{the increment in acceleration}}$ $$x(2t)-x(t)=v(t)\cdot t+ \frac12t^2(f\color{red}{+f})\ \text { meters}$$ and will have a final velocity of $$v(2t)=v(t)+t(f+f)\ \frac{m}{s}.$$ It is now quite clear, this you can prove by inspection, that between time $kt$ and time $(k+1)t$ your object will travel $$x((k+1)t)-x(kt)=v(kt)\cdot t+ \frac12t^2((k+1)f)\ \text { meters}$$ and its final velocity will be $$v((k+1)t)-v(kt)=t((k+1)f)\ \frac{m}{s}.$$ Now we sum all contributions, to see that $$v(Kt)=\sum_{k=0}^{K-1}\left[v((k+1)t)-v(kt)\right]+v(0)=\sum_{k=0}^{K-1}\left[t((k+1)f)\right]+0=tf \sum_{k=0}^{K-1}(k+1)=tf\frac{K(K+1)}{2},$$ so that in particular, $$v(nt)=tf\frac{n(n+1)}{2}.$$ Similarly \begin{align*}x(nt)& =\sum_{k=0}^{n-1}\left[x((k+1)t)-x(kt)\right]+x(0)=\sum_{k=0}^{n-1}\left[v(kt)\cdot t+ \frac12t^2((k+1)f)\right]+x(0)\\ & =t^2f\sum_{k=0}^{n-1}\frac{k(k+1)}{2}+ \frac12t^2 f\sum_{k=0}^{n-1} (k+1)\\& = t^2f\sum_{k=0}^{n-1}\left[\frac{k(k+1)}{2}+\frac{k+1}{2}\right] \\ & = \frac{t^2f}{2}\sum_{k=0}^{n-1}(k+1)^2 \\ & = \frac{t^2f}{2}\cdot\frac{n(n+1)(2n+1)}{6} \end{align*}
Where in the last step I have used the sum of squares formula (How to get to the formula for the sum of squares of first n numbers?).
Remark: I suppose the suggested answer is missing a $1/6$ factor
