Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Question
Watch Video
Show Examples
The position of an object moving in a straight line, in kilometers, can be modeled by the function 
s(t), where 
t is measured in days. The velocity of the object is 2 kilometers per day when 
t=5. Selected values of 
s(t) are shown in the table below. Use a linear approximation when 
t=5 to estimate the position of the object at time 
t=4.8. Use proper units.





t
0
5
8
13
18
20



s(t)
30
40
55
80
90
100

Question\newlineWatch Video\newlineShow Examples\newlineThe position of an object moving in a straight line, in kilometers, can be modeled by the function s(t) s(t) , where t t is measured in days. The velocity of the object is 22 kilometers per day when t=5 t=5 . Selected values of s(t) s(t) are shown in the table below. Use a linear approximation when t=5 t=5 to estimate the position of the object at time t=4.8 t=4.8 . Use proper units.\newline\begin{tabular}{|c|c|c|c|c|c|c|}\newline\hlinet t & 00 & 55 & 88 & 1313 & 1818 & 2020 \\\newline\hlines(t) s(t) & 3030 & 4040 & 5555 & 8080 & 9090 & 100100 \\\newline\hline\newline\end{tabular}

Full solution

Q. Question\newlineWatch Video\newlineShow Examples\newlineThe position of an object moving in a straight line, in kilometers, can be modeled by the function s(t) s(t) , where t t is measured in days. The velocity of the object is 22 kilometers per day when t=5 t=5 . Selected values of s(t) s(t) are shown in the table below. Use a linear approximation when t=5 t=5 to estimate the position of the object at time t=4.8 t=4.8 . Use proper units.\newline\begin{tabular}{|c|c|c|c|c|c|c|}\newline\hlinet t & 00 & 55 & 88 & 1313 & 1818 & 2020 \\\newline\hlines(t) s(t) & 3030 & 4040 & 5555 & 8080 & 9090 & 100100 \\\newline\hline\newline\end{tabular}
  1. Identify Given Values: Step 11: Identify the given values and the requirement.\newlineWe know the velocity at t=5t=5 is 22 km/day and we need to estimate the position at t=4.8t=4.8 using linear approximation. We have s(5)=40s(5) = 40 km from the table.
  2. Calculate Change in Time: Step 22: Calculate the change in time.\newlineThe change in time, Δt\Delta t, from t=5t=5 to t=4.8t=4.8 is Δt=4.85=0.2\Delta t = 4.8 - 5 = -0.2 days.
  3. Use Velocity for Change in Position: Step 33: Use the velocity to find the change in position.\newlineSince the velocity at t=5t=5 is 22 km/day, the change in position, Δs\Delta s, can be approximated by Δs=velocity×Δt=2\Delta s = \text{velocity} \times \Delta t = 2 km/day ×0.2\times -0.2 days =0.4= -0.4 km.
  4. Estimate Position at t=4.8t=4.8: Step 44: Estimate the position at t=4.8t=4.8. Using the change in position, estimate s(4.8)=s(5)+Δs=40km0.4km=39.6kms(4.8) = s(5) + \Delta s = 40 \, \text{km} - 0.4 \, \text{km} = 39.6 \, \text{km}.

More problems from Solve quadratic equations: word problems