Question

an object travels along a straight line. the function $s(t) = \sqrt{2t + 1}$ gives the object’s position, in meters, at time $t$ hours. what is the object’s acceleration, in meters per second per second, at $t = 1.5$? write your answer as an integer or decimal rounded to the nearest tenth. meters per second per second

Explanation:

Step1: Recall velocity and acceleration

Velocity \( v(t) \) is the first derivative of position \( s(t) \), and acceleration \( a(t) \) is the second derivative of \( s(t) \) (or first derivative of \( v(t) \)). Given \( s(t)=\sqrt{2t + 1}=(2t + 1)^{\frac{1}{2}} \).

Step2: Find first derivative (velocity)

Using the power rule \( \frac{d}{dt}(u^n)=nu^{n - 1}u' \), where \( u = 2t+1 \), \( n=\frac{1}{2} \), \( u' = 2 \).
\( v(t)=s'(t)=\frac{1}{2}(2t + 1)^{-\frac{1}{2}}\times2=(2t + 1)^{-\frac{1}{2}} \)

Step3: Find second derivative (acceleration)

Differentiate \( v(t)=(2t + 1)^{-\frac{1}{2}} \) again. Using power rule: \( n=-\frac{1}{2} \), \( u = 2t + 1 \), \( u' = 2 \).
\( a(t)=v'(t)=-\frac{1}{2}(2t + 1)^{-\frac{3}{2}}\times2=-(2t + 1)^{-\frac{3}{2}} \)

Step4: Evaluate at \( t = 1.5 \)

First, convert \( t = 1.5 \) hours to seconds? Wait, no, wait the problem says position is in meters at time \( t \) hours? Wait, no, maybe it's a typo, or maybe \( t \) is in seconds? Wait, no, the acceleration is in meters per second per second, so probably \( t \) is in seconds? Wait, the problem says "at time \( t \) hours" but acceleration is m/s², so maybe it's a mistake, but proceed with \( t = 1.5 \) (assuming \( t \) is in the appropriate unit for the function, maybe the function has \( t \) in seconds? Wait, no, let's check the function \( s(t)=\sqrt{2t + 1} \). Let's compute \( 2t+1 \) when \( t = 1.5 \): \( 2(1.5)+1=4 \). Then \( a(1.5)=-(4)^{-\frac{3}{2}}=-\frac{1}{4^{\frac{3}{2}}}=-\frac{1}{(2^2)^{\frac{3}{2}}}=-\frac{1}{2^3}=-\frac{1}{8}=-0.125 \). Wait, but the problem says "meters per second per second", so maybe the time unit in the function is seconds. Wait, but the problem says "time \( t \) hours" – that's confusing. But assuming that maybe it's a mistake and \( t \) is in seconds, or maybe the function is correct as is. Wait, let's re - check the differentiation:

Wait, \( s(t)=\sqrt{2t + 1} \), so \( s'(t)=\frac{1}{\sqrt{2t + 1}} \), \( s''(t)=-\frac{2}{2(2t + 1)^{\frac{3}{2}}}=-\frac{1}{(2t + 1)^{\frac{3}{2}}} \). When \( t = 1.5 \), \( 2t+1=4 \), so \( s''(1.5)=-\frac{1}{4^{\frac{3}{2}}}=-\frac{1}{8}=-0.125 \). So the acceleration is \( - 0.125\space m/s^2 \), or rounded to nearest tenth is \( - 0.1\space m/s^2 \)? Wait, no, \( - 0.125 \) rounded to nearest tenth is \( - 0.1 \)? Wait, no, 0.125 rounded to nearest tenth: the hundredth digit is 2, which is less than 5? Wait, no, 0.125: tenths place is 1, hundredths is 2, thousandths is 5. Wait, when rounding to nearest tenth, look at hundredths place. 0.125: hundredths digit is 2, which is less than 5, so we keep tenths digit as 1? Wait, no, 0.125 is 0.1 when rounded to nearest tenth? Wait, no, 0.125 is closer to 0.1 or 0.2? The distance from 0.1 to 0.125 is 0.025, from 0.125 to 0.2 is 0.075. So it's closer to 0.1. But wait, maybe I made a mistake in differentiation.

Wait, let's re - do the differentiation:

\( s(t)=(2t + 1)^{\frac{1}{2}} \)

First derivative: \( s'(t)=\frac{1}{2}(2t + 1)^{-\frac{1}{2}}\times2=(2t + 1)^{-\frac{1}{2}} \) (correct)

Second derivative: \( s''(t)=-\frac{1}{2}(2t + 1)^{-\frac{3}{2}}\times2=-(2t + 1)^{-\frac{3}{2}} \) (correct)

At \( t = 1.5 \), \( 2t+1=2\times1.5 + 1=4 \)

So \( s''(1.5)=-(4)^{-\frac{3}{2}}=-\frac{1}{4^{\frac{3}{2}}}=-\frac{1}{(4^{\frac{1}{2}})^3}=-\frac{1}{2^3}=-\frac{1}{8}=-0.125 \)

So the acceleration is \( - 0.125\space m/s^2 \), which is \( - 0.1\space m/s^2 \) when rounded to the nearest tenth. Wait, but maybe the time unit was a mistake, and \( t \) is in seconds. Alternatively, maybe the problem has a typo and \( t \) is in seconds. So the acceleration at \( t = 1.5 \) (assuming \( t \) is in seconds) is \( - 0.1\space m/s^2 \) (rounded to nearest tenth) or \( - 0.125\space m/s^2 \) as a decimal.

Wait, but the problem says "Write your answer as an integer or decimal rounded to the nearest tenth." So \( - 0.125\) rounded to nearest tenth is \( - 0.1 \)? Wait, no, 0.125: the tenths place is 1, hundredths is 2, so we round down, so \( 0.1 \). But with the negative sign, it's \( - 0.1 \). Wait, but let's check again.

Wait, maybe I messed up the derivative. Let's use another method. Let's compute the first derivative at \( t = 1.5 \) and \( t = 1.5+h \), then take the limit as \( h\to0 \) for acceleration (but that's more complicated). But the power rule is correct. So the second derivative is \( -(2t + 1)^{-\frac{3}{2}} \), at \( t = 1.5 \), it's \( - 4^{-\frac{3}{2}}=- \frac{1}{8}=-0.125 \), which is \( - 0.1 \) when rounded to the nearest tenth (since the hundredth digit is 2, which is less than 5, so we keep the tenth digit as 1).

Answer:

\(-0.1\) (or \(-0.125\) if not rounded, but the problem says round to nearest tenth, so \(-0.1\))