1) A particle moves in a straight line whth initial velocity -2 m / ec. from a point 3 m . in the positive direction from a fixed point on the straight line such that $$ a=21+1 $$ , then $$ x= $$ $$ \qquad $$ m . when the speed has vanished. (a) $$ \frac{11}{6} $$ (b) $$ \frac{11}{3} $$ (c) $$ \frac{-2}{3} $$ (d) 1 2)If $$ v=3 t^{2}-2 t, x=1 $$ when $$ t=0 $$, then $$ \qquad $$ (a) $$ x=6 t-2 $$ (b) $$ x=3 t^{2}-2 t+1 $$ (c) $$ x=1^{3}-t^{2}+1 $$ (d) $$ x=t^{3}-t^{2}-1 $$ 3)If $$ \mathrm{v}=1+\sin \mathrm{t}, x=-3 $$ when $$ \mathrm{t}=0 $$, then (a) $$ x=t+\cos t $$ (b) $$ x=t-\cos t $$ (C) $$ x=t-\cos t+2 $$ (d) $$ x=t-\cos t-2 $$ 4)If $$ v(t)=9.8 \mathrm{t}+5 $$ where $$ x(0)=10 $$, then $$ x(10)= $$ (a) zero (b) 530 (C) 540 (d) 550 5)Velocity $$ v=t(2-t) \mathrm{m} / \mathrm{sec} $$. and $$ X(0)=-3 $$, then $$ X(6)= $$ (a) -105 (b) -39 (C) 39 (d) 105

Question

UpStudy AI · Accepted Answer

1) Given the acceleration  
$$
a(t)=2t+1,
$$
and the initial velocity  
$$
v(0)=-2,
$$
we first find the velocity by integrating the acceleration:
$$
v(t)=\int a(t)\,dt=\int (2t+1)\,dt=t^2+t+C.
$$
Using $$v(0)=-2$$, we have:
$$
v(0)=0+0+C=-2\quad\Longrightarrow\quad C=-2.
$$
Thus,
$$
v(t)=t^2+t-2.
$$

The speed vanishes when $$v(t)=0$$. Solve:
$$
t^2+t-2=0.
$$
Factorizing,
$$
(t+2)(t-1)=0.
$$
Since time must be positive, we take $$t=1$$.

Next, the position is found by integrating the velocity:
$$
x(t)=\int v(t)\,dt=\int (t^2+t-2)\,dt=\frac{t^3}{3}+\frac{t^2}{2}-2t+C.
$$
The initial condition is given as $$x(0)=3$$. Thus,
$$
x(0)=0+0-0+C=3\quad\Longrightarrow\quad C=3.
$$
So,
$$
x(t)=\frac{t^3}{3}+\frac{t^2}{2}-2t+3.
$$

Evaluating at $$t=1$$:
$$
x(1)=\frac{1}{3}+\frac{1}{2}-2+3.
$$
Compute step by step:
$$
\frac{1}{3}+\frac{1}{2}=\frac{2}{6}+\frac{3}{6}=\frac{5}{6},
$$
so
$$
x(1)=\frac{5}{6}-2+3=\frac{5}{6}+1=\frac{5}{6}+\frac{6}{6}=\frac{11}{6}.
$$

Thus, the answer is  
$$
\boxed{\frac{11}{6}}.
$$

2) We are given 
$$
v=3t^2-2t,\quad \text{and}\quad x(0)=1.
$$
Integrate the velocity to get the position:
$$
x(t)=\int v(t)\,dt=\int (3t^2-2t)\,dt=t^3-t^2+C.
$$
Use the initial condition:
$$
x(0)=0-0+C=1\quad\Longrightarrow\quad C=1.
$$
Thus:
$$
x(t)=t^3-t^2+1.
$$
The answer is  
$$
\boxed{t^3-t^2+1}.
$$

3) We have 
$$
v=1+\sin t,\quad \text{and}\quad x(0)=-3.
$$
Integrate the velocity:
$$
x(t)=\int (1+\sin t)\,dt=t-\cos t+C.
$$
Using $$x(0)=-3$$:
$$
x(0)=0-\cos0+C=-1+C=-3\quad\Longrightarrow\quad C=-2.
$$
So:
$$
x(t)=t-\cos t-2.
$$
Thus, the answer is  
$$
\boxed{t-\cos t-2}.
$$

4) Given:
$$
v(t)=9.8t+5\quad \text{and}\quad x(0)=10.
$$
Integrate $$v(t)$$ to determine $$x(t)$$:
$$
x(t)=\int (9.8t+5)\,dt=4.9t^2+5t+C.
$$
Using $$x(0)=10$$:
$$
x(0)=C=10.
$$
Thus:
$$
x(t)=4.9t^2+5t+10.
$$
Evaluating at $$t=10$$:
$$
x(10)=4.9(10)^2+5(10)+10=4.9(100)+50+10=490+50+10=550.
$$
Hence, the answer is  
$$
\boxed{550}.
$$

5) The particle’s velocity is given by:
$$
v=t(2-t)=2t-t^2,\quad \text{and}\quad X(0)=-3.
$$
Integrate the velocity:
$$
X(t)=\int (2t-t^2)\,dt=t^2-\frac{t^3}{3}+C.
$$
Using $$X(0)=-3$$:
$$
X(0)=C=-3.
$$
Thus,
$$
X(t)=t^2-\frac{t^3}{3}-3.
$$
Now evaluate at $$t=6$$:
$$
X(6)=6^2-\frac{6^3}{3}-3=36-\frac{216}{3}-3=36-72-3=-39.
$$
So, the answer is  
$$
\boxed{-39}.
$$
1. $$ x = \frac{11}{6} $$ meters when the speed has vanished. 2. $$ x = t^3 - t^2 + 1 $$ 3. $$ x = t - \cos t - 2 $$ 4. $$ x(10) = 550 $$ 5. $$ X(6) = -39 $$

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Calculus Questions