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The position function of a moving object is \( s(t)=t^{\frac{5}{2}}(7-t), t \geq...
Mar 7, 2024
The position function of a moving object is s(t)=t52(7−t),t≥0 s(t)=t^{\frac{5}{2}}(7-t), t \geq 0 , in metres, at time t t , in seconds. show steps a) Calculate the object's velocity and acceleration at any time t t . b) After how many seconds does the object stop ? c) When does the motion of the object change direction ? d) When is its acceleration positive ? e) When does the object return to its original position ?
Generated Graph
Solution by Steps
step 1
To find the velocity, we differentiate the position function s(t)=t52(7−t) s(t) = t^{\frac{5}{2}}(7-t)
step 2
Using the product rule, ddt[t52(7−t)]=ddt[t52]⋅(7−t)+t52⋅ddt[−t] \frac{d}{dt}[t^{\frac{5}{2}}(7-t)] = \frac{d}{dt}[t^{\frac{5}{2}}] \cdot (7-t) + t^{\frac{5}{2}} \cdot \frac{d}{dt}[-t]
step 3
Differentiating t52 t^{\frac{5}{2}} and −t -t gives 52t32 \frac{5}{2}t^{\frac{3}{2}} and −1 -1 respectively
step 4
Substituting the derivatives, we get v(t)=52t32(7−t)−t52 v(t) = \frac{5}{2}t^{\frac{3}{2}}(7-t) - t^{\frac{5}{2}}
step 5
Simplifying, v(t)=352t32−72t52−t52 v(t) = \frac{35}{2}t^{\frac{3}{2}} - \frac{7}{2}t^{\frac{5}{2}} - t^{\frac{5}{2}}
step 6
Combining like terms, v(t)=352t32−92t52 v(t) = \frac{35}{2}t^{\frac{3}{2}} - \frac{9}{2}t^{\frac{5}{2}}
step 7
To find the acceleration, we differentiate the velocity function v(t) v(t)
step 8
Differentiating v(t) v(t) gives a(t)=ddt[352t32]−ddt[92t52] a(t) = \frac{d}{dt}[\frac{35}{2}t^{\frac{3}{2}}] - \frac{d}{dt}[\frac{9}{2}t^{\frac{5}{2}}]
step 9
Differentiating t32 t^{\frac{3}{2}} and t52 t^{\frac{5}{2}} gives 32⋅352t12 \frac{3}{2} \cdot \frac{35}{2}t^{\frac{1}{2}} and 52⋅92t32 \frac{5}{2} \cdot \frac{9}{2}t^{\frac{3}{2}} respectively
step 10
Substituting the derivatives, we get a(t)=1054t12−454t32 a(t) = \frac{105}{4}t^{\frac{1}{2}} - \frac{45}{4}t^{\frac{3}{2}}
Answer
Velocity: v(t)=352t32−92t52 v(t) = \frac{35}{2}t^{\frac{3}{2}} - \frac{9}{2}t^{\frac{5}{2}} , Acceleration: a(t)=1054t12−454t32 a(t) = \frac{105}{4}t^{\frac{1}{2}} - \frac{45}{4}t^{\frac{3}{2}}
Key Concept
Differentiation of power functions using the power rule and the product rule.
Explanation
The velocity is the first derivative of the position function, and the acceleration is the derivative of the velocity function.
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Solution by Steps
step 1
The object stops when its velocity is zero
step 2
Set the velocity function v(t)=352t32−92t52 v(t) = \frac{35}{2}t^{\frac{3}{2}} - \frac{9}{2}t^{\frac{5}{2}} equal to zero
step 3
Solve 352t32−92t52=0 \frac{35}{2}t^{\frac{3}{2}} - \frac{9}{2}t^{\frac{5}{2}} = 0 for t t
step 4
Factor out t32 t^{\frac{3}{2}} , getting t32(352−92t)=0 t^{\frac{3}{2}}(\frac{35}{2} - \frac{9}{2}t) = 0
step 5
Set each factor equal to zero, t32=0 t^{\frac{3}{2}} = 0 and 352−92t=0 \frac{35}{2} - \frac{9}{2}t = 0
step 6
Solving 352−92t=0 \frac{35}{2} - \frac{9}{2}t = 0 gives t=359 t = \frac{35}{9}
Answer
The object stops after t=359 t = \frac{35}{9} seconds.
Key Concept
Setting the velocity function equal to zero to find when the object stops.
Explanation
The object's velocity is zero when t=359 t = \frac{35}{9} seconds, which is when the object stops moving.
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Solution by Steps
step 1
The object changes direction when the velocity changes sign
step 2
Find the values of t t where v(t)=352t32−92t52 v(t) = \frac{35}{2}t^{\frac{3}{2}} - \frac{9}{2}t^{\frac{5}{2}} is zero or undefined
step 3
We already found that v(t)=0 v(t) = 0 at t=359 t = \frac{35}{9} seconds
step 4
The velocity is undefined at t=0 t = 0 , but since t≥0 t \geq 0 , we only consider t=359 t = \frac{35}{9}
Answer
The object changes direction at t=359 t = \frac{35}{9} seconds.
Key Concept
The change in direction occurs when the velocity function changes sign.
Explanation
The velocity function changes sign at t=359 t = \frac{35}{9} seconds, indicating a change in direction.
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Solution by Steps
step 1
The acceleration is positive when a(t) = \frac{105}{4}t^{\frac{1}{2}} - \frac{45}{4}t^{\frac{3}{2}} > 0
step 2
Solve \frac{105}{4}t^{\frac{1}{2}} - \frac{45}{4}t^{\frac{3}{2}} > 0 for t t
step 3
Factor out t12 t^{\frac{1}{2}} , getting t^{\frac{1}{2}}(\frac{105}{4} - \frac{45}{4}t) > 0
step 4
Set the factors equal to zero, t12=0 t^{\frac{1}{2}} = 0 and 1054−454t=0 \frac{105}{4} - \frac{45}{4}t = 0
step 5
Solving 1054−454t=0 \frac{105}{4} - \frac{45}{4}t = 0 gives t=10545 t = \frac{105}{45}
step 6
The acceleration is positive for 0 < t < \frac{105}{45}
Answer
The acceleration is positive for 0 < t < \frac{105}{45} seconds.
Key Concept
Determining when the acceleration function is greater than zero.
Explanation
The acceleration is positive when t t is between 0 and 10545 \frac{105}{45} seconds.
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Solution by Steps
step 1
The object returns to its original position when s(t)=t52(7−t)=0 s(t) = t^{\frac{5}{2}}(7-t) = 0
step 2
Solve t52(7−t)=0 t^{\frac{5}{2}}(7-t) = 0 for t t
step 3
Set each factor equal to zero, t52=0 t^{\frac{5}{2}} = 0 and 7−t=0 7-t = 0
step 4
Solving 7−t=0 7-t = 0 gives t=7 t = 7
Answer
The object returns to its original position at t=0 t = 0 and t=7 t = 7 seconds.
Key Concept
Setting the position function equal to zero to find when the object returns to its original position.
Explanation
The object is at its original position at the start t=0 t = 0 and when t=7 t = 7 seconds.
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