Physics, 18.11.2020 04:10 mandilynn22

An engineer measures the velocity of a remote-controlled cart on a straight track at regular time intervals. The data are shown in the graph above. During which of the following time intervals did the cart return to its position at time t=0 s? A. 7 s ≤ t < 10 s
B. 10 s ≤ t < 12 s
C. 5 s ≤ t < 6 s
D. 3 s ≤ t < 5 s

An engineer measures the velocity of a remote-controlled cart on a straight track at regular time i

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Answer from: zabattle

The car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.

The change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:

A + A' = 0 (1)

A is the change in position from t = 0 and t = 3 s, measured in metres.

A' is the change in position from t = 5 and t = T s, measured in metres.

As per the triangle law,

1/2 ×(3 s)(1.2) - 1/2 ×(T - 5)v = 0

1.8 - 0.5T(v) +2.5(v) =0

T = (1.8 + 2.5v)/ (0.5v)

Here, v is the final velocity of cart.

For v = 1 m/s, we have,

T = (1.8 + 2.5(1))/ (0.5(1))

T = 8.6 s

Since, the values do not coincide with the time from the graph, so taking another value, v = 1.5 m/s to obtain the time as,

T = (1.8 + 2.5(1.5))/ (0.5(1.5))

T = 7.4 s

This value does not coincide with the time from the graph. If we know that , v = 1.1 m/s, then the intended instant is:

T = (1.8 + 2.5(1.1))/ (0.5(1.1))

T = 8.273 s

Thus, we can conclude that the car will return to its position at the time interval of 7 s ≤ t < 10 s. Hence, option (A) is correct.

Learn more about the speed-time graph here:

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Answer from: lex3340

A. 7 s ≤ t < 10 s

Explanation:

From Definite Integral Theory and Mechanical Physics we remember that change in position is represented by sum of areas below the curve. According to the graphic, we see that the following condition must be satisfied:

$A_{1}+A_{2} = 0$ (1)

Where:

$A_{1}$ - Change in position from $t = 0\,s$ and $t = 3\,s$ , measured in meters.

$A_{2}$ - Change in position from $t = 5\,s$ and t = T , measured in meters.

By definitions of triangle, we expand the equation above:

$\frac{1}{2}\cdot (3\,s)\cdot \left(1.2\,\frac{m}{s} \right) -\frac{1}{2}\cdot (T-5\,s)\cdot v = 0\,m$

$1.8\,m-0.5\cdot T \cdot v +2.5\cdot v = 0$

And the time is now cleared:

$1.8\,m+2.5\cdot v = 0.5\cdot T\cdot v$

$T = \frac{1.8+2.5\cdot v}{0.5\cdot v}$

Where:

- Final velocity of the cart, measured in meters per second.

- Time, measured in seconds.

If we know that $v = 1\,\frac{m}{s}$ , then the intended instant is:

$T = \frac{1.8+2.5\cdot (1)}{0.5\cdot (1)}$

$T = 8.6\,s$

The value does not coincide with the time from the graph.

If we know that $v = 1.5\,\frac{m}{s}$ , then the intended instant is:

$T = \frac{1.8+2.5\cdot (1.5)}{0.5\cdot (1.5)}$

$T = 7.4\,s$

This value does not coincide with the time from the graph.

If we know that $v = 1.1\,\frac{m}{s}$ , then the intended instant is:

$T = \frac{1.8+2.5\cdot (1.1)}{0.5\cdot (1.1)}$

$T = 8.273\,s$

This results offer a reasonable approximation, which that the correct answer is A.

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