An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 6.00 s. Its speed as it passes the second point is 15.0 m/s. What are (a) its speed at the first point and (b) its acceleration?

Joseph Krupa
2021-12-14
Answered

An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 6.00 s. Its speed as it passes the second point is 15.0 m/s. What are (a) its speed at the first point and (b) its acceleration?

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Getting acceleration due to gravity from dropping ball experiment

This seems like pretty basic experiment, but I'm having a lot of trouble with it. Basically, I have two timer gates that measure time between two signals, and I drop metal ball between them. This way I'm getting distance traveled, and time. Ball is dropped from right above the first gate to make sure initial velocity is as small as possible (no way to make it 0 with this setup/timer). I'm assuming $v$ initial is $0\frac{m}{s}$. Gates are $1$ meter apart.

Times are pretty consistent, and average result from dropping ball from $1.0$ meters is $0.4003$ seconds.

So now I have $3$ [constant acceleration] equations that I can use to get $g$

1.

${d}_{traveled}={v}_{initial}.t+\frac{1}{2}a{t}^{2}$

$a=\frac{2d}{{t}^{2}}$

$a=\frac{2.1}{(.4003{)}^{2}}$

$a=12.48\frac{m}{{s}^{2}}$

2.

${v}_{f}^{2}={v}_{i}^{2}+2ad$

$a=\frac{{v}_{f}^{2}-{v}_{i}^{2}}{2d}$

${v}_{f}=\frac{distance}{time}=\frac{1.0}{0.4003}=2.5\frac{m}{s}$

$a=\frac{(2.5\frac{m}{s}{)}^{2}}{2.1m}$

$a=3.125\frac{m}{{s}^{2}}$

3.

${v}_{f}={v}_{i}+at$

$a=\frac{{v}_{f}-{v}_{i}}{t}$

$a=\frac{2.5m/s-0}{0.4003s}$

$a=6.25\frac{m}{{s}^{2}}$

I'm getting three different results. And all of them are far from $9.8\frac{m}{{s}^{2}}$ . No idea what I'm doing wrong.

Also, if I would drop that ball from different heights, and plot distance-time graph, how can I get acceleration from that?

This seems like pretty basic experiment, but I'm having a lot of trouble with it. Basically, I have two timer gates that measure time between two signals, and I drop metal ball between them. This way I'm getting distance traveled, and time. Ball is dropped from right above the first gate to make sure initial velocity is as small as possible (no way to make it 0 with this setup/timer). I'm assuming $v$ initial is $0\frac{m}{s}$. Gates are $1$ meter apart.

Times are pretty consistent, and average result from dropping ball from $1.0$ meters is $0.4003$ seconds.

So now I have $3$ [constant acceleration] equations that I can use to get $g$

1.

${d}_{traveled}={v}_{initial}.t+\frac{1}{2}a{t}^{2}$

$a=\frac{2d}{{t}^{2}}$

$a=\frac{2.1}{(.4003{)}^{2}}$

$a=12.48\frac{m}{{s}^{2}}$

2.

${v}_{f}^{2}={v}_{i}^{2}+2ad$

$a=\frac{{v}_{f}^{2}-{v}_{i}^{2}}{2d}$

${v}_{f}=\frac{distance}{time}=\frac{1.0}{0.4003}=2.5\frac{m}{s}$

$a=\frac{(2.5\frac{m}{s}{)}^{2}}{2.1m}$

$a=3.125\frac{m}{{s}^{2}}$

3.

${v}_{f}={v}_{i}+at$

$a=\frac{{v}_{f}-{v}_{i}}{t}$

$a=\frac{2.5m/s-0}{0.4003s}$

$a=6.25\frac{m}{{s}^{2}}$

I'm getting three different results. And all of them are far from $9.8\frac{m}{{s}^{2}}$ . No idea what I'm doing wrong.

Also, if I would drop that ball from different heights, and plot distance-time graph, how can I get acceleration from that?

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(a) Find the acceleration of the elevator (magnitude and direction).

(b) What is the acceleration if the scale reads 670 N?

(c) If the scale reads zero, should the student worry? Explain.

(d) What is the tension in the cable in parts (a) and (с)?