
How do you find values of t in which the speed of the particle is increasing if the position of a particle moving along a line is given by \[s\left( t \right) = 2{t^3} - 24{t^2} + 90t + 7\] for \[t \geqslant 0\]?
Answer
539.7k+ views
Hint: To you find values of t in which the speed of the particle is increasing if the position of a particle moving along a line, we need to find the derivative of the given speed i.e., \[s\left( t \right) = 2{t^3} - 24{t^2} + 90t + 7\] and then find its second derivative and then consider some values of time t and find the speed; then compare it with respect to t.
Complete step by step answer:
Let us write the given data:
If,
\[s\left( t \right) = 2{t^3} - 24{t^2} + 90t + 7\] ………………… 1
The speed at time t is given by finding its derivative, and we know that \[\dfrac{d}{{dx}}{t^3} = 3{t^2}\] and \[\dfrac{d}{{dx}}{t^2} = 2t\], hence applying this to the equation 1 we get:
\[s\left( t \right) = 2\left( {3{t^2}} \right) - 24\left( {2{t^2}} \right) + 90\left( 1 \right)\]
\[ \Rightarrow s'\left( t \right) = 6{t^2} - 48t + 90\]
Acceleration is given by finding its second derivative as:
\[s^{''}\left( t \right) = 6\left( {2t} \right) - 48\left( 1 \right)\]
\[ \Rightarrow s^{''}\left( t \right) = 12t - 48\]
The critical point for acceleration (when the speed changes from decreasing to increasing) occurs when,
\[s^{''}\left( t \right) = 12t - 48 = 0\]
That is when, \[t = 4\]
This might be easier to see if we consider a table of time(t) and speed at time t:
Distance\[\left( t \right) = 2{t^3} - 24{t^2} + 90t + 7\]
Speed\[\left( t \right) = 6{t^2} - 48t + 90\]
We can see that the speed actually decreases until somewhere between \[t = 3\] and \[t = 5\] when it starts to increase again.
Note: The key point while solving this sum is that, you must know all the basics of derivatives, such that in the given terms, we need to find the second derivative also, hence to solve these kinds of questions, you must know the derivative rules of all the functions and the derivative rules include the constant rule, power rule, constant multiple rules, sum rule, and difference rule.
Complete step by step answer:
Let us write the given data:
If,
\[s\left( t \right) = 2{t^3} - 24{t^2} + 90t + 7\] ………………… 1
The speed at time t is given by finding its derivative, and we know that \[\dfrac{d}{{dx}}{t^3} = 3{t^2}\] and \[\dfrac{d}{{dx}}{t^2} = 2t\], hence applying this to the equation 1 we get:
\[s\left( t \right) = 2\left( {3{t^2}} \right) - 24\left( {2{t^2}} \right) + 90\left( 1 \right)\]
\[ \Rightarrow s'\left( t \right) = 6{t^2} - 48t + 90\]
Acceleration is given by finding its second derivative as:
\[s^{''}\left( t \right) = 6\left( {2t} \right) - 48\left( 1 \right)\]
\[ \Rightarrow s^{''}\left( t \right) = 12t - 48\]
The critical point for acceleration (when the speed changes from decreasing to increasing) occurs when,
\[s^{''}\left( t \right) = 12t - 48 = 0\]
That is when, \[t = 4\]
This might be easier to see if we consider a table of time(t) and speed at time t:
Distance\[\left( t \right) = 2{t^3} - 24{t^2} + 90t + 7\]
Speed\[\left( t \right) = 6{t^2} - 48t + 90\]
| t | Speed(t) |
| 0 | 90 |
| 1 | 48 |
| 2 | 18 |
| 3 | 0 |
| 4 | -6 |
| 5 | 0 |
| 6 | 18 |
We can see that the speed actually decreases until somewhere between \[t = 3\] and \[t = 5\] when it starts to increase again.
Note: The key point while solving this sum is that, you must know all the basics of derivatives, such that in the given terms, we need to find the second derivative also, hence to solve these kinds of questions, you must know the derivative rules of all the functions and the derivative rules include the constant rule, power rule, constant multiple rules, sum rule, and difference rule.
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