What is its velocity? What is its acceleration?

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What is its velocity? What is its acceleration?

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Answer 1 · 2018-02-10T10:49:55

$x (2.5) = 3.435$ $x(2.5)=3.435$
$v (2.5) = - 1.91$ $v(2.5)=-1.91$
$a (t) = - 2$ $a(t)=-2$

Explanation:

It is given that the displacement $x$ $x$ at time $t$ $t$ is given by $x (t) = 1.96 + 3.09 t - t^{2}$ $x(t)=1.96+3.09t-t^2$ .

At $t = 2.5$ $t=2.5$ , the displacement $x (2.5) = 1.96 + 3.09 \cdot 2.5 - {2.5}^{2} = 3.435$ $x(2.5)=1.96+3.09*2.5-2.5^2=3.435$ .

The velocity $v$ $v$ is the derivative of displacement $x$ $x$ with respect to time $t$ $t$ , or $v (t) = \frac{d x}{d t} = 3.09 - 2 t$ $v(t)=dx/dt=3.09-2t$ . Thus, the velocity at time $t = 2.5$ $t=2.5$ is $v (2.5) = 3.09 - 2 \cdot 2.5 = - 1.91$ $v(2.5)=3.09-2*2.5=-1.91$ .

The acceleration $a$ $a$ is the derivative of velocity $v$ $v$ with respect to time $t$ $t$ , or $a (t) = \frac{d v}{d t} = - 2$ $a(t)=(dv)/dt=-2$ . Since it stays constant regardless of the time, the acceleration at time $t = 2.5$ $t=2.5$ is $- 2$ $-2$ .

Answer 2 · 2018-02-10T10:52:22

The particle moves according to the equation, $1.96 + 3.09 t - t^{2}$ $1.96 + 3.09t - t^2$
so,
$x = 1.96 + 3.09 t - t^{2}$ $x=1.96 + 3.09t - t^2$
Velocity is the rate of change of displacement, $x$ $x$
therefore, $v = \frac{d x}{d t}$ $v=dx/dt$

$v = \frac{d (1.96 + 3.09 t - t^{2})}{d t}$ $v=(d(1.96 + 3.09t - t^2))/dt$

$v = (3.09 - 2 t) \frac{m}{s}$ $v=(3.09-2t)m/s$

Now, acceleration is the rate of change of velocity, $v$ $v$
therefore, $a = d \frac{v}{d t}$ $a=dv/dt$

$a = \frac{d (3.09 - 2 t)}{d t}$ $a=(d(3.09-2t))/dt$

$a = - 2$ $a=-2$ $\frac{m}{s^{2}}$ $m/s^2$

Answer 3 · 2018-02-10T10:52:32

Given, $x = 1.96 + 3.09 t - t^{2}$ $x=1.96+3.09t-t^2$

at, $t = 2.50$ $t=2.50$ , $s = 1.96 + (3.09 \cdot 2.50) - {(2.50)}^{2} = 3.435 m$ $s=1.96+(3.09*2.50)-(2.50)^2=3.435 m$

differentiating the equation w.r.t time,we get,

velocity = $\frac{d x}{d t} = 3.09 - 2 t$ $(dx)/(dt) = 3.09 -2t$

So,at $t = 2.50 s$ $t=2.50s$ its velocity is $3.09 - (2 \cdot 2.50) = - 1.91 \frac{m}{s}$ $3.09 -(2*2.50)=-1.91 m/s$

And,differentiating it one again we get, acceleration= $\frac{d v}{d t} = - 2 \frac{m}{s^{2}}$ $(dv)/(dt) = -2 m/s^2$

ALTERNATIVELY

You can compare the equation with, $s = u t - \frac{1}{2} a t^{2}$ $s=ut -1/2 at^2$

So,you can rearrange as, $x - 1.96 = 3.09 t - \frac{1}{2} \times 2 \times t^{2}$ $x-1.96=3.09t - 1/2 ×2×t^2$

So comparing we can say,it had an initial velocity of $3.09 \frac{m}{s}$ $3.09 m/s$ (at $t = 0$ $t=0$ ) acceleration is $- 2 \frac{m}{s^{2}}$ $-2 m/s^2$ and at $t = 0$ $t=0$ it was $1.96 m$ $1.96 m$ to the left of origin i.e along negative $X$ $X$ axis.

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What is its velocity? What is its acceleration?

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