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.

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