Question

A train starting from rest, travels with uniform acceleration and acquires a velocity of 80 $\frac{km}{h}$ in two minute. Find the acceleration and the distance traveled by train during this time.

Answer 1

Step 1: Given that:

Initial velocity (u) = 0 .

Final velocity $$\begin{gathered} V=80\frac{km}{h}\times \frac{5}{18}=\frac{400}{18}\frac{m}{s}. \\ V=22.22\frac{m}{s}. \end{gathered}$$

Time (t) =2 minutes =120 seconds .

Step 2: Formula

$v=u+at$

$s=ut+\frac{1}{2}\times a\times t^2.$

Step 3:Calculation

Acceleration $\left(a\right)=\frac{v-u}{t}.$

$$\begin{gathered} a=\frac{(22.22-0){\displaystyle \frac{m}{s}}}{120s}. \\ a=\frac{22.22}{120}. \\ a=0.185\frac{m}{s^2}. \end{gathered}$$

Distance

$s=ut+\frac{1}{2}\times a\times t^2.$

$$\begin{gathered} s=0\times 120+\frac{1}{2}\times 0.185\times 120\times 120. \\ s=0.185\times 60\times 120. \\ s=1332m \end{gathered}$$

Step 4: Final Answer:

Hence,

Acceleration (a) =$0.185\frac{m}{s^2}.$

Distance (s) = 1332 m.