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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

Ask by Hall Vaughn. in Egypt
Mar 15,2025

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**14)** The body stops after 4 seconds, and the distance from point \(O\) is 32 cm. **15)** After 6 seconds, the distance between the two bodies is 216 meters. **16)** - After 2 seconds, the body is at 11 cm from \(O\). - The displacement from 1 to 5 seconds is 0 cm. - The total distance traveled from 1 to 5 seconds is 8 cm. **17)** At \(x=14\) meters, the speed of the particle is 24 m/sec.

Solution

**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.

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Initially, using the formula of motion with constant acceleration, we can find the distance the body covers before it stops. Its acceleration, given as \( a = (12 - 6t) \text{ cm/sec}^2 \), will eventually result in a zero velocity, indicating the distance can be calculated by integrating the acceleration function until the velocity becomes zero, arriving at the correct option of 64 cm. Now, for the next question about two bodies moving from the same point, we can compute their respective positions at 6 seconds. The first body has a non-linear velocity equation, creating a uniquely different position from the constant velocity of the second body. By integrating the velocity of the first body from 0 to 6 seconds and subtracting the second body's distance, we reveal that the distance between them is indeed 108 meters.

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