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3 years ago

Do you add the initial position and final position to get distance?

1 answer:

4 0

You can do d=st (s is speed and t is time) or you can use pythagorean theorem if the distance is or the two positions is a right angle.

You might be interested in

What is the frequency of a wave that has a wave speed of 80m/s and a wavelength of 0.20m

IrinaK [193]

speed=frequency*wavelength, so frequency=speed/wavelength. frequency=80*0.2

Frequency = 16 Hz

6 0

3 years ago

A 3-kg skateboard is rolling down the sidewalk at 4 m/s when it collides with a 1-kg skateboard that was initially at rest. If t

irga5000 [103]

Answer:

2 m/s

Explanation:

Parameters given:

Mass of first skateboard, m = 3 kg

Initial speed of first skateboard, u = 4 m/s

Mass of second skateboard, M = 1 kg

Initial speed of second skateboard, U = 0 m/s

Final speed of second skateboard, V = 6 m/s

Using the principle of the conservaton of momentum, the total initial momentum is equal to the total final momentum.

Momentum is the product of mass and velocity. This implies that:

m*u + M*U = m*v + M*V

(3*4) + (1*0) = (3*v) + (1*6)

12 + 0 = 3v + 6

=> 3v = 12 - 6

3v = 6

v = 6/3 = 2 m/s

The final speed of the 3 kg skateboard is 2 m/s

8 0

4 years ago

Read 2 more answers

A force F= (6.70 N)i + (5.60 N)j + (6.60 N)k act on a 9.80 kg mobile object that moves from an initial position of di = (9.10 m)

Mandarinka [93]

Answer:

W = 0.11J

P = 0.0186W

Explanation:

Please see attachment below.

4 0

3 years ago

A car is moving with speed 20 m/s and acceleration 2 m/s2 at a given instant. Using a second-degree Taylor polynomial, estimate

PilotLPTM [1.2K]

Answer:

$T(1)=21$

Explanation:

The equation of the position in kinematics is given:

$x(t)=x_{0}+v_{0}t+0.5at^{2}$

x(0) is the initial position, in this it is 0
v(0) is the initial velocity (20 m/s)
a is the acceleration (2 m/s²)

So the equation will be:

$x(t)=20t+0.5*2*t^{2}$

$x(t)=20t+t^{2}$

Now, the Taylor polynomial equation is:

$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+...$

Using our position equation we can find f'(t)=v(t) and f''(x)=a(t). In our case a=0, so let's find each derivative.

$f(t)=x(t)=20t+t^{2}$

$f'(t)=\frac{dx(t)}{dt}=v(t)=20+2t$

$f''(t)=\frac{dv(t)}{dt}=a(t)=2$

Using the Taylor polynomial with a = 0 and take just the second order of the derivative.

$f(0)+\frac{f'(0)}{1!}(x)+\frac{f''(0)}{2!}(x)^{2}$

$f(0)=x(0)=0$

$f'(0)=v(0)=20$

$f''(0)=a(0)=2$

$T(t)=f(0)+\frac{f'(0)}{1!}(t)+\frac{f''(0)}{2!}(t)^{2}$

$T(t)=\frac{20}{1!}(t)+\frac{2}{2!}(t)^{2}$

$T(t)=20t+t^{2}$

Let's put t=1 so find the how far the car moves in the next second:

$T(1)=20*1+1^{2}$

$T(1)=21$

Therefore, the position in the next second is 21 m.

We need to know if the acceleration remains at this value to use this polynomial in the next minute, so I suggest that it would be reasonable to use this method just under this condition.

I hope it helps you!

4 0

3 years ago

Most sodium atoms have an atomic mass of 23, but a few have atomic

irina1246 [14]

Answer:

Explanation:

8 0

3 years ago