Physics, 09.11.2020 16:40 sergun252005

The car on a track begins at the 80-cm mark. The car moves in the negative direction at a speed of 5 cm/s. After three seconds, a. What distance has the car traveled? b. What is the displacement of the car? c. What is the car's new position on the track? Please hurry up

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Answer from: ccollingwood

a) The car travelled a distance of 15 centimeters.

b) The displacement of the car is -15 centimeters.

c) The new position of the car on the track is 65 centimeters.

Explanation:

a) From Mechanical Physics we know that the distance travelled is magnitude of the route covered by the vehicle. If this car travels in a straight line and at constant velocity, then we can calculate the distance traveled by the following expression:

$\Delta s = v\cdot \Delta t$ (Eq. 1)

Where:

- Speed of the car, measured in centimeters per second.

$\Delta t$ - Time, measured in seconds.

$\Delta s$ - Distance travelled by the car, measured in meters.

If we get that $v = 5\,\frac{cm}{s}$ and $\Delta t = 3\,s$ , then the distance travelled by the car is:

$\Delta s = \left(5\,\frac{cm}{s} \right)\cdot (3\,s)$

$\Delta s = 15\,cm$

The car travelled a distance of 15 centimeters.

b) The displacement is the vectorial distance between final and initial position. Then, we find that initial and final positions of the car are, respectively:

Initial position:

$s_{o} = 80\,cm$

Final position: ( $s_{o} = 80\,cm$ , $\Delta s = 15\,cm$ )

$s_{f} = s_{o} -\Delta s$ (Eq. 1)

$s_{f} = 80\,cm-15\,cm$

$s_{f} = 65\,cm$

And the displacement of the car is: ( $s_{o} = 80\,cm$ , $s_{f} = 65\,cm$ )

$d = s_{f}-s_{o}$ (Eq. 2)

$d = 65\,cm-80\,cm$

$d = -15\,cm$

The displacement of the car is -15 centimeters.

c) The position of the car is its location regarding a point of reference. From point b) we find that the new position of the car on the track is:

$s_{f} = 65\,cm$

The new position of the car on the track is 65 centimeters.

Answer from: Quest

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.

neglecting air resistance etc etc.,

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.

vh = 8 ms-1

the vertical velocity component has an accelerating force: 'g'.

using v = u +at where a = g = 9.81 ms-2.

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.

adding vectorally gives: -

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

Answer from: Quest

c.the charges if the proton and electron and gravity

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