14) A body starts its motion from rest from a fixed point \( (O) \) in a horizontal straight line tvith acceleration \( a=(12-61) \mathrm{cm} / \mathrm{sec}^{2} \), where the time \( t \) in seconds , so the distance betiveen the body and the point \( (O) \) when it stops instantly \( = \). \( \qquad \) cm. (a) 16 (b) 32 (c) 64 (d) 96 15) Tivo bodies began their motions in a straight line from the same point and at the same time and in the same direction, one of them moves with velocity \( v_{1}=\left(3 t^{2}+2 t\right) \mathrm{m} / \mathrm{sec} \), and the other moves ivith velocity \( v_{2}=(21) \mathrm{m} / \mathrm{sec} \)., then the distance berween them after 6 seconds from the beginning \( = \) \( \qquad \) m. (a) 6 (b) 36 (C) 108 (d) 216 16) Aody moves in a straight line with velocity \( \mathrm{v}=(6-2 \mathrm{t}) \mathrm{cm} \). \( / \mathrm{sec} \)., the body began its motion 3 cm . on the right side of a fixed point \( (O) \) on the straight line then : First: The position of the body after 2 seconds from the beginning of motion = \( \qquad \) cm. (a) 8 (b) 11 (c) 12 (d) 19 Second : The displacement in the time interval \( [1,5] \) equals (a) -10 (b) zero (C) 10 (d) 20 Third: The total distance in the time interval \( [1,5] \) equals (a) zero (b) 4 (c) 8 (d) 12 17) A particle moves in a straight line from a fixed point (O) with initial speed \( 10 \mathrm{~m} / \mathrm{sec} \). such that the algebraic measure of its acceleration is given in term of the algebraic measure of its position s by the relation \( a=2 x+3 \), then its speed at \( x=14 \mathrm{~m} \), equals \( \qquad \) \( \mathrm{m} / \mathrm{sec} \). (a) 24 (b) 34 (c) 476 (d) 576

**14)** The acceleration is given by \[ a(t)=12-6t\quad \text{(in cm/sec}^2\text{)}. \] Since the body starts from rest, the velocity is \[ v(t)=\int_0^t a(\tau)\,d\tau=\int_0^t (12-6\tau)d\tau=12t-3t^2. \] The body stops when \(v(t)=0\). Thus, \[ 12t-3t^2=0\quad\Longrightarrow\quad 3t(4-t)=0. \] Discarding \(t=0\) (the initial time), we have \(t=4\) seconds. The displacement is obtained from \[ s(t)=\int_0^t v(\tau)\,d\tau=\int_0^t(12\tau-3\tau^2)d\tau. \] Evaluating the integral at \(t=4\): \[ s(4)=\left[6\tau^2-\tau^3\right]_0^4=6(16)-64=96-64=32\quad \text{cm}. \] Answer: 32 cm. --- **15)** For the first body, the velocity is given by \[ v_1(t)=3t^2+2t. \] Its displacement (starting from 0) is \[ s_1(t)=\int_0^t (3\tau^2+2\tau)d\tau=\left[\tau^3+\tau^2\right]_0^t=t^3+t^2. \] At \(t=6\) seconds: \[ s_1(6)=6^3+6^2=216+36=252\quad \text{m}. \] For the second body, assuming its velocity is given by \[ v_2(t)=2t\quad \text{(in m/sec)}, \] its displacement is \[ s_2(t)=\int_0^t 2\tau\,d\tau=\left[\tau^2\right]_0^t=t^2. \] At \(t=6\) seconds: \[ s_2(6)=6^2=36\quad \text{m}. \] The distance between them after 6 seconds is \[ 252-36=216\quad \text{m}. \] Answer: 216 m. --- **16)** The velocity is \[ v(t)=6-2t\quad \text{(in cm/sec)}, \] and the initial position is 3 cm (to the right of \(O\)). *First: Position after 2 seconds* The position function is \[ x(t)=3+\int_0^t (6-2\tau)d\tau. \] Evaluating the integral: \[ \int_0^2 (6-2\tau)d\tau=\left[6\tau-\tau^2\right]_0^2=6(2)-2^2=12-4=8. \] Thus, \[ x(2)=3+8=11\quad \text{cm}. \] *Second: Displacement in the time interval \([1,5]\)* Displacement is given by \(x(5)-x(1)\). First compute \(x(t)\) in general: \[ x(t)=3+\left(6t-t^2\right). \] Then, \[ x(5)=3+(30-25)=3+5=8,\quad x(1)=3+(6-1)=3+5=8. \] So, displacement \(=8-8=0\quad \text{cm}\). *Third: Total distance in \([1,5]\)* First, determine if the velocity changes sign. Set \[ 6-2t=0\quad\Longrightarrow\quad t=3. \] Between \(t=1\) and \(t=3\), \(v(t)>0\); between \(t=3\) and \(t=5\), \(v(t)<0\). Distance traveled from \(t=1\) to \(t=3\): \[ D_1=\int_1^3 (6-2t)d t. \] Compute: \[ \int_1^3 (6-2t)d t=\left[6t-t^2\right]_1^3=\big[(18-9)-(6-1)\big]=(9)-(5)=4. \] Distance traveled from \(t=3\) to \(t=5\): Since \(6-2t\) is negative, take absolute value: \[ D_2=\int_3^5 (2t-6)d t. \] Compute: \[ \int_3^5 (2t-6)d t=\left[t^2-6t\right]_3^5=\Big[(25-30)-(9-18)\Big]=(-5)-(-9)=4. \] Total distance \(=D_1+D_2=4+4=8\quad \text{cm}\). Answers: Position after 2 sec: 11 cm; Displacement from 1 to 5 sec: 0 cm; Total distance from 1 to 5 sec: 8 cm. --- **17)** The acceleration is given in terms of position by \[ a=2x+3. \] Recall that \[ a=v\frac{dv}{dx}. \] Thus, \[ v\frac{dv}{dx}=2x+3. \] Separate variables: \[ v\,dv=(2x+3)\,dx. \] Integrate both sides. The initial condition is at \(x=0\) with \(v=10\) m/sec. For a general position \(x\) (with corresponding speed \(v\)): \[ \int_{10}^{v} v\,dv=\int_{0}^{x}(2x+3)\,dx. \] The left side becomes: \[ \left[\frac{v^2}{2}\right]_{10}^{v}=\frac{v^2-10^2}{2}. \] The right side is: \[ \int_0^x (2x+3)dx =\left[x^2+3x\right]_0^x=x^2+3x. \] Thus, \[ \frac{v^2-100}{2}=x^2+3x. \] We are to find \(v\) at \(x=14\) m. Substitute \(x=14\): \[ \frac{v^2-100}{2}=14^2+3(14)=196+42=238. \] Multiply both sides by 2: \[ v^2-100=476, \] \[ v^2=576. \] Taking the positive square root (since speed is positive): \[ v=24\quad \text{m/sec}. \] Answer: 24 m/sec.