Super Sleigh On Frictionless Ice

Since , initially the sledge is at rest , therefore

$\boxed{u = 0}$

also, we know that :

$\boxed{v^{2} - u^{2} = 2as}$

where s = distance travelled and v = final velocity.

So, just plug in the values : u = 0 , s = 9 m , and a = 2 m/ $s^{2}$

we get v = $\sqrt{2as}$ = 6 m/s

We use the equation of motion v^2 = u^2 + 2as where v is its resultant velocity, u is its initial velocity, a is the acceleration and s is the displacement with respect to starting point. Clearly u = 0, a = 2, s = 9. Substituting these values into the equation, we get v = 6.

Much like any other motion problem, let us begin by finding the equation of motion that governs this particular situation. Unsurprisingly, Newton´s second law is used:

$\Sigma$ $\overrightarrow{F}$ = $\frac{\mathrm d}{\mathrm d t}$ $\overrightarrow{P}$

where $\overrightarrow{P}$ is the linear momentum vector and $\overrightarrow{F}$ the force vector. Seeing as we may assume that the mass of the sled remains constant, we can simplify the above vector equation to the form:

$\Sigma$ $\overrightarrow{F}$ = $m\frac{\mathrm d}{\mathrm d t}$ $\overrightarrow{V}$

since the derivative of the mass with respect to time is zero and recalling that:

$\overrightarrow{P}$ = $m\overrightarrow{V}$

With the push, the equation of motion must account for a net force acting on the sled:

$2m$ = $m\frac{\mathrm d}{\mathrm d t}$ $V$

We now have a scalar equation since the respective force vector and velocity vector components were set equal to each other. Of course an inertial frame of reference is used to describe these vectors. Solving this simple differential equation gives:

$\int$ $2dt$ = $\int$ $dV$

$V(t)$ = $2t+k$

Assuming the sled is at rest initially:

$V(t)$ = $2t$

Now we know how the velocity of the sled varies with time. The next step is to determine how its position varies with time. To do this, we integrate the above equation again with respect to time knowing that the derivative of position with time is velocity we have:

$2t$ = $\frac{\mathrm d}{\mathrm d t}$ $X$

$\int$ $2tdt$ = $\int$ $dX$

$X(t)$ = $t^2+k$

Considering the position at zero seconds to be zero:

$X(t)$ = $t^2$

We now know how the sled´s position varies with time. Knowing that the sled moved $9$ meters, we can solve for the time taken for this to happen:

$t^2$ = $9$ $\Rightarrow$ $t = 3 seconds$

Finally, using this information in our velocity function we obtain the solution:

$V(t=3)$ = $2(3)$ = $\boxed{6 m/s}$

By using newton's eqn of motion,

v^2-u^2=2as v^2=229 = 36 therefore, v=6 m/s.

v^2-u^2=2 a s; where u=0,s=9,a=2 therefore, by equating v get v=6ms^-1

by using laws of kinematics.. v2=u2+2as..u=0,a=2,s=9.....hence,v2=2 9 2=36, that implies v=6

as we know, V^2 =v^2+2 a x d V = Velocity after traveling 9 meters; v = the first velocity; a = the acceleration; d = is the distance.

V^2 = 0 + 2 x 2 x 9 V^2 = 36 V = 6 m/s

Using third equ: of motion.... 2aS = vf^2 - vi^2 vi = 0, vf^2 = 2 2 9 = 36 vf = 6 m/s

Given: a=2 m/s^2 d=9 m Vi=0 m/s

Required: Vf

Solution: 2ad = Vf^2 - Vi^2

                    2(2 m/s^2)(9 m) = Vf^2 - (0 m/s)^2

                    Vf^2 = 36 m^2/s^2

                    Vf = 6 m/s     ]ans.

the given acceleration is 2 m/s^2. As the question says that the sleigh rests on a smooth frictionless sheet of ice, therefore u = 0. The distance travelled is 9m. Therefore, by using newton's eqn of motion,

v^2-u^2=2as v^2=2 2 9 = 36 therefore, v=6 m/s.

u=0 v=? a=2m/s sq s=9m

by third eq. of motion . v(sq) - u(sq) = 2as by substituting the values , we get v= 6 , answer

Vt squared = Vo squared + 2.a.x Vt squared = 0 + 2 . 2 . 9 Vt squared = 36 Vt = 36 rooted Vt = 6

v2=v02+2ad v2=0+2.2.9 entao: v=6m/s

s = Vo.t + 1/2 . a . t^2 = 0 + 1/2 . a . t^2 = 1/2 . a . t^2 ---> 9 = 1/2 . 2 . t^2 ----> t = 3. So, Vt = Vo + at ---> Vt = 0 + 2 . 3 [m/s] = 6 m/s. Answer : 6

Let the initial and final velocity after travelling 9 m be 'u' and 'v' resp,. Let the distance travelled be s $=9$ m. Also, let the acceleration of the sleigh be a $=2$ m/ $s^{2}$ . Since, sleigh starts from rest, we have u $=0$ .

We here use the motion equation $v^{2}=u^{2}+2as$ to solve this problem.

So, $v^{2}=u^{2}+2as \implies v^{2}=0^{2}+2\times 2\times 9 \implies v^{2}=0+36 \implies v^{2}=36 \implies v=6 [$ Neglecting $(-6)$ as velocity cannot be negative. $]$

So, we finally get $v=\boxed{6}$

We use the equation $v^2=v_0^2+2a\Delta x$ . Plugging in, we have $v^2 = 2(2)(9) = 36$ , and $v=\pm 6$ . Assuming the direction of push to be positive, we reject the negative solution and have the answer as $x=6$ .