The velocity of a car travelling on a straigh | vedclass

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(B) Initial velocity $u = 3.6 \ km/h = 3.6 	imes \frac{5}{18} \ m/s = 1 \ m/s$.Final velocity $v = 2 	imes u = 2 \ m/s$.Time $t = 10 \ s$.Accelerati

Vedclass · Accepted Answer

$95$

Vedclass · Answer

(B) Initial velocity $u = 3.6 \ km/h = 3.6 	imes \frac{5}{18} \ m/s = 1 \ m/s$.Final velocity $v = 2 	imes u = 2 \ m/s$.Time $t = 10 \ s$.Acceleration $a = \frac{v - u}{t} = \frac{2 - 1}{10} = 0.1 \ m/s^2$.Distance covered $s = ut + \frac{1}{2}at^2 = (1)(10) + \frac{1}{2}(0.1)(10)^2 = 10 + 5 = 15 \ m$.Circumference of the wheel $C = 2 \pi r = 2 \pi (0.25) = 0.5 \pi \ m$.Number of revolutions $n = \frac{s}{C} = \frac{15}{0.5 \pi} = \frac{30}{\pi} \approx \frac{30}{3.14159} \approx 9.54$.Wait, checking the calculation: $n = 9.54$ revolutions. Re-evaluating the question context: If the question implies angular displacement $	heta$ in radians, $	heta = \frac{s}{r} = \frac{15}{0.25} = 60 \ rad$. Number of revolutions $n = \frac{	heta}{2 \pi} = \frac{60}{2 \pi} = \frac{30}{\pi} \approx 9.55$. Given the options, there might be a unit or scale factor discrepancy in the problem statement, but based on standard kinematics, the value is approximately $9.5$. If the intended answer is $95$, it implies a factor of $10$ difference.

Column-$I$	Column-$II$
$(1)$ Velocity-time relation	$(a)$ $v = v_0 + at$
$(2)$ Velocity-displacement relation	$(b)$ $S = v_0t + \frac{1}{2}at^2$
	$(c)$ $v^2 = v_0^2 + 2as$

Similar Questions

Match the relations in Column-$I$ with the equations in Column-$II$ for uniformly accelerated motion.

Column-$I$ Column-$II$

$(1)$ Velocity-time relation $(a)$ $v = v_0 + at$

$(2)$ Velocity-displacement relation $(b)$ $S = v_0t + \frac{1}{2}at^2$

$(c)$ $v^2 = v_0^2 + 2as$

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