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

Question

UpStudy AI · Accepted Answer

**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.
**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.

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions