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r-ruslan [8.4K]
3 years ago
10

If the plotted points on a speed-time graph do not form a straight line, what do you know about the object's acceleration?

Physics
2 answers:
slega [8]3 years ago
6 0
Remember, that while sped is constant, acceleration is not. Acceleration is when velicity changes. So the graph which shows the slop <span>of a velocity vs time describes acceleration.
</span>If we have the straight line on the graph it means that the slope is always the same whereas the <span>non-linear graphs has a variable slope that changes depending on your point in the graph.
</span>To conclude - if your graph is not a straight line it has variable acc at many points.<span>

</span>
Morgarella [4.7K]3 years ago
3 0

I accidentally answered the wrong one. Im sorry (edit)

Explanation:

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A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/
navik [9.2K]

To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

\text{Initial Momentum} = \text{Final Momentum}

m_1u_1 +m_2u_2 = m_1v_1+m_2v_2

Here,

m_{1,2}= Mass of each object

u_{1,2}= Initial velocity of each object

v_{1,2}= Final velocity of each object

When they position the final velocities of the bodies it is the same and the car is stationary then,

m_2u_2 = (m_1+m_2)v_f

Rearranging to find the final velocity

v_f = \frac{m_2u_2}{ (m_1+m_2)}

v_f = \frac{ 12500*7.8}{ 12500+7430}

v_f = 4.8921ft/s

The expression for the impulse received by the first car is

I = m_1 (v-u)

I = \frac{W}{g} (v-u)

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The negative sign show the opposite direction.

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