Question

an object falls because of gravity at a rate of -980 cm/s² (- sign is indicative of a downward direction) for 10. seconds. if it starts from rest (v₀ = 0) and its position starts from d₀= 0.0, where is it 10. seconds later? m -49 -4900 -490

Explanation:

Step1: Recall the kinematic equation for displacement

The kinematic equation for displacement \( d \) when an object has constant acceleration \( a \), initial velocity \( v_0 \), initial position \( d_0 \), and time \( t \) is \( d = d_0 + v_0t+\frac{1}{2}at^2 \).
Given \( d_0 = 0.0 \, \text{cm} \), \( v_0 = 0 \, \text{cm/s} \), \( a=- 980 \, \text{cm/s}^2 \), and \( t = 10 \, \text{s} \).

Step2: Substitute the values into the equation

Substitute the values into the formula:
\( d=0 + 0\times10+\frac{1}{2}\times(-980)\times(10)^2 \)
First, calculate \( (10)^2=100 \). Then, \( \frac{1}{2}\times(-980)\times100=- 490\times100=-49000 \, \text{cm} \).

Step3: Convert centimeters to meters

Since \( 1 \, \text{m}=100 \, \text{cm} \), to convert \( - 49000 \, \text{cm} \) to meters, we divide by 100: \( \frac{-49000}{100}=-490 \, \text{m} \)? Wait, no, wait. Wait, \( \frac{1}{2}\times(-980)\times10^2=\frac{1}{2}\times(-980)\times100=-490\times100 = - 49000 \, \text{cm} \). Convert to meters: \( -49000\div100=-490 \, \text{m} \)? Wait, no, wait, \( 980 \, \text{cm/s}^2=9.8 \, \text{m/s}^2 \), maybe I made a mistake in unit conversion earlier. Wait, let's re - do the unit conversion properly.

Wait, the acceleration is \( a=-980 \, \text{cm/s}^2 \). Let's keep the units consistent. Let's work in centimeters first and then convert to meters.

\( d = d_0+v_0t+\frac{1}{2}at^2 \)
\( d_0 = 0 \, \text{cm} \), \( v_0 = 0 \, \text{cm/s} \), \( a=-980 \, \text{cm/s}^2 \), \( t = 10 \, \text{s} \)

\( d=0 + 0\times10+\frac{1}{2}\times(-980)\times(10)^2 \)
\(=\frac{1}{2}\times(-980)\times100=-490\times100=-49000 \, \text{cm} \)

Now, convert \( \text{cm} \) to \( \text{m} \): since \( 1 \, \text{m} = 100 \, \text{cm} \), \( d=\frac{-49000}{100}=-490 \, \text{m} \)? Wait, no, wait, \( 980 \, \text{cm/s}^2=9.8 \, \text{m/s}^2 \), let's use \( a = - 9.8 \, \text{m/s}^2 \) (since \( 980 \, \text{cm/s}^2=\frac{980}{100}\text{m/s}^2 = 9.8 \, \text{m/s}^2 \)). Then \( t = 10 \, \text{s} \), \( v_0 = 0 \, \text{m/s} \), \( d_0 = 0 \, \text{m} \)

\( d=d_0 + v_0t+\frac{1}{2}at^2=0+0\times10+\frac{1}{2}\times(-9.8)\times(10)^2 \)
\(=\frac{1}{2}\times(-9.8)\times100=-4.9\times100=-490 \, \text{m} \)? Wait, no, \( \frac{1}{2}\times(-9.8)\times100=-490 \)? Wait, \( 9.8\times50 = 490 \), so with the negative sign, it's \( - 490 \, \text{m} \)? Wait, but earlier when we used \( \text{cm} \), we had \( - 49000 \, \text{cm}=-490 \, \text{m} \) (since \( 49000\div100 = 490 \)). So the displacement is \( - 490 \, \text{m} \). Wait, but let's check the options. The options are - 49, - 4900, - 490.

Wait, maybe I made a mistake in the acceleration unit. Wait, the problem says the acceleration is \( - 980 \, \text{cm/s}^2 \). Let's recalculate with \( \text{cm} \) and then convert to meters correctly.

\( d=\frac{1}{2}at^2 \) (since \( d_0 = 0 \), \( v_0 = 0 \))

\( a=-980 \, \text{cm/s}^2 \), \( t = 10 \, \text{s} \)

\( d=\frac{1}{2}\times(-980)\times(10)^2=\frac{1}{2}\times(-980)\times100=-490\times100=-49000 \, \text{cm} \)

Now, convert \( \text{cm} \) to \( \text{m} \): \( 1 \, \text{m}=100 \, \text{cm} \), so \( d=\frac{-49000}{100}=-490 \, \text{m} \). Wait, but the options have - 490. Wait, maybe the acceleration was supposed to be \( - 980 \, \text{cm/s}^2=-9.8 \, \text{m/s}^2 \), and \( t = 10 \, \text{s} \)

\( d=\frac{1}{2}\times(-9.8)\times(10)^2=\frac{1}{2}\times(-9.8)\times100=-490 \, \text{m} \). So the correct answer is - 490.

Answer:

-490