Uniformly Accelerated Motion Questions in English

(D) Let the total time of travel be $T$ seconds and acceleration be $a$. The initial velocity $u = 0$.
Distance covered in the first second ($t=1$ s) is given by $s_1 = ut + \frac{1}{2}at^2$.
$1 = 0(1) + \frac{1}{2}a(1)^2 \Rightarrow a = 2 \,m/s^2$.
Distance covered in the last second is given by the formula for distance in the $n^{th}$ second: $s_n = u + \frac{a}{2}(2n - 1)$.
Here, $s_n = 39 \,m$, $u = 0$, $a = 2 \,m/s^2$.
$39 = 0 + \frac{2}{2}(2T - 1) \Rightarrow 39 = 2T - 1 \Rightarrow 2T = 40 \Rightarrow T = 20 \,s$.
The total distance $S$ covered in $T = 20 \,s$ is given by $S = ut + \frac{1}{2}aT^2$.
$S = 0(20) + \frac{1}{2}(2)(20)^2 = 400 \,m$.

(A) Given, initial velocity $u = 54 \,km/h = 54 \times \frac{5}{18} \,m/s = 15 \,m/s$.
Distance of signal from vehicle $d = 400 \,m$, acceleration $a = -0.3 \,m/s^2$.
When the vehicle stops, final velocity $v = 0$.
Using the equation of motion $v^2 = u^2 + 2as$:
$0^2 = (15)^2 + 2(-0.3)s$
$0 = 225 - 0.6s$
$0.6s = 225$
$s = \frac{225}{0.6} = 375 \,m$.
The vehicle's position relative to the traffic light is $d - s = 400 \,m - 375 \,m = 25 \,m$.

(B) Using the first equation of motion,$v = u + at$. Since both cars start from rest $(u = 0)$,their velocities after time $t$ are:
$v_1 = a_1 t$ and $v_2 = a_2 t$
Subtracting these,we get: $(v_1 - v_2) = (a_1 - a_2)t \dots (1)$
Using the second equation of motion,$s = ut + \frac{1}{2}at^2$. The distance between them is:
$s_1 - s_2 = \frac{1}{2}a_1 t^2 - \frac{1}{2}a_2 t^2 = \frac{1}{2}(a_1 - a_2)t^2$
Given $s_1 - s_2 = 50 \ m$ and $(a_1 - a_2) = 4 \ m \ s^{-2}$,we have:
$50 = \frac{1}{2} \times 4 \times t^2 \Rightarrow 50 = 2t^2 \Rightarrow t^2 = 25 \Rightarrow t = 5 \ s$
Substituting $t = 5 \ s$ and $(a_1 - a_2) = 4 \ m \ s^{-2}$ into equation $(1)$:
$(v_1 - v_2) = 4 \times 5 = 20 \ m \ s^{-1}$.

(B) Initial velocity $u = 3.6 \ km/h = 3.6 \times \frac{5}{18} \ m/s = 1 \ m/s$.
Final velocity $v = 2 \times u = 2 \ m/s$.
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 $\theta$ in radians,$\theta = \frac{s}{r} = \frac{15}{0.25} = 60 \ rad$. Number of revolutions $n = \frac{\theta}{2 \pi} = \frac{60}{2 \pi} = \frac{30}{\pi} \approx 9.55$. 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.