Question

A train starting from a railway station and moving with uniform acceleration attains a speed $40km{h^{ - 1}}$ in 10 minutes. Find its acceleration.
$A.$ $0.0185\dfrac{m}{{{s^2}}}$
$B.$ $0.0195\dfrac{m}{{{s^2}}}$
$C.$ $0.0165\dfrac{m}{{{s^2}}}$
$D.$ $0.0135\dfrac{m}{{{s^2}}}$

Answer 1

HINT- In mechanics, acceleration is the rate of change of velocity of an object with respect to time. Accelerations are vector quantities (in that they have magnitude and direction). The orientation of an object’s acceleration is given by the orientation of the net force acting on that object. The magnitude of an object’s acceleration, as described by Newton’s second law of motion, is the combined effect of two causes:
$1 -$ the net balance of all external forces acting on that object- magnitude is directly proportional to this net resulting force;
$2 -$ that object’s mass, depending on the materials out of which it is made- magnitude is inversely proportional to the object’s mass.
The SI unit for acceleration is metre per second ($\dfrac{m}{{{s^2}}}$ )

Complete step-by-step answer:
As the train started from the railway station, so initially it must be in the rest. Hence, the initially velocity of the train will be $u = 0$
As the train attains the speed of $40\dfrac{{km}}{{{h^{}}}}$ in 10 minutes, therefore, the final velocity of the train is given by
$v = 40\dfrac{{km}}{h} \Rightarrow 40 \times \dfrac{5}{{18}} = 11.11\dfrac{m}{\operatorname{s} }$ (we converted $\dfrac{{km}}{h} \Rightarrow \dfrac{m}{s}$ by multiplying it by $\dfrac{5}{{18}}$ )
Now, $t = 10\min = 10 \times 60 = 600\sec$
By using first law of motion,
$v = u + at$
$11.11 = 0 + a(600)$
$a = \dfrac{{11.11}}{{600}}$
$a = 0.0185\dfrac{m}{{{s^2}}}$ , which gives option $A.$ as the correct option.

NOTE- Uniform or constant acceleration is a type of motion in which the velocity of an object changes by an equal amount changes by an equal amount in equal time period. A frequently cited example of uniform acceleration is that of an object in free fall in a uniform gravitational field strength g (also called acceleration due to gravity). By Newton’s second law the force (${F_g}$ ) acting on the body is given by:
${F_g} = mg$