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PART - 1 CH - 2 Motion in a Straight Line — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsPART - 1 CH - 2 Motion in a Straight Line3 Marks
Question
A train start moving from rest and its speed increases with uniform acceleration for 10 sec . At the end of 5 sec . its speed become $54 km /$ hour.
(a) Find the acceleration of the train.
(b) What will be its speed after 10 sec ?
(c) How much distance will the train travel in 10 sec?
(d) How much distance did the train cover in the seventh and tenth second respectively?
✓

Answer

(A) speed after $5 sec .=54 km / hr$.
$=54 \times \frac{5}{18} m / s=15 m / s$
At rest $u =0, v =15 m / s , t =5 sec , a =$ ?
From the first equation of motion,
$v=u+at$
$\begin{array}{ll}
\therefore \quad a & =\frac{v-u}{t}=\frac{15-0}{5} \\
& a=3 m / s^2
\end{array}$
(b) Velocity after $10 sec . v = u +$ at
$=0+3 \times 10=30 m / s$
(c) Distance
$\begin{aligned}
s & =u t+\frac{1}{2} at^2 \\
& =0 \times 10+\frac{1}{2} \times 3 \times 10 \times 10 \\
& =150 \text { meter }
\end{aligned}$
(d) Distance covered in nth second
$s_{n}=u+\frac{1}{2} a(2 n-1)$
Therefore distance in 7th second
$\begin{aligned}
s_7 & =0+\frac{1}{2} \times 3(2 \times 7-1) \\
& =\frac{39}{20}=19.5 \text { meter }
\end{aligned}$
Distance in 10th sec.
$\begin{aligned}
s_{10} & =0+\frac{1}{2} \times 3 \times(2 \times 10-1) \\
& =\frac{57}{2}=28.5 \text { meter }
\end{aligned}$

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