
The distance covered by a particle varies with time as \[x = \dfrac{k}{b}\left( {1 -
{e^{ - bt}}} \right)\]. The speed of the particle at time t is
A. \[k\,{e^{ - bt}}\]
B. \[kb\,{e^{ - bt}}\]
C. \[\left( {\dfrac{k}{{{b^2}}}} \right)\,{e^{ - bt}}\]
D. \[\left( {\dfrac{k}{b}} \right)\,{e^{ - bt}}\]
Answer
565.8k+ views
Hint:The speed of the particle at the instant of time t can be obtained by differentiating the expression for the distance covered by the particle in time t. Therefore, differentiate the expression for the distance covered by the particle with respect to time t to get the expression for the speed of the particle.
Formula used:
Speed,\[v = \dfrac{{dx}}{{dt}}\]
Here, x is the distance covered and t is the time.
Complete step by step answer:
We have given the distance covered by the particle with time t as,
\[x = \dfrac{k}{b}\left( {1 - {e^{ - bt}}} \right)\] …… (1)
Here, t is the time.
We have the speed of the particle is the first derivative of the displacement relation with
respect time. Therefore,
\[v = \dfrac{{dx}}{{dt}}\]
Here, x is the distance covered and t is the time.
Using equation (1) in the above equation, we get,
\[v = \dfrac{d}{{dt}}\left( {\dfrac{k}{b}\left( {1 - {e^{ - bt}}} \right)} \right)\]
\[ \Rightarrow v = \dfrac{k}{b}\dfrac{d}{{dt}}\left( {1 - {e^{ - bt}}} \right)\]
\[ \Rightarrow v = \dfrac{k}{b}\left( { - {e^{ - bt}}\left( { - b} \right)} \right)\]
\[ \Rightarrow v = k\,{e^{ - bt}}\]
Here, k and b are the constant quantities.
The above expression gives the speed of the particle at time t.
So, the correct answer is option (A).
Note: Be careful while differentiating the exponential terms. The derivative of the term
\[\dfrac{d}{{dx}}\left( {{e^{nx}}} \right) = n{e^{nx}}\] and not just \[{e^{nx}}\]. The expression
for the speed that we have derived only gives the speed of the particle at the instant of time
t and not the average velocity of the particle after time t. If you differentiate the expression
for the speed with respect to time t, you will get the acceleration of the particle at that
instant of time.
Formula used:
Speed,\[v = \dfrac{{dx}}{{dt}}\]
Here, x is the distance covered and t is the time.
Complete step by step answer:
We have given the distance covered by the particle with time t as,
\[x = \dfrac{k}{b}\left( {1 - {e^{ - bt}}} \right)\] …… (1)
Here, t is the time.
We have the speed of the particle is the first derivative of the displacement relation with
respect time. Therefore,
\[v = \dfrac{{dx}}{{dt}}\]
Here, x is the distance covered and t is the time.
Using equation (1) in the above equation, we get,
\[v = \dfrac{d}{{dt}}\left( {\dfrac{k}{b}\left( {1 - {e^{ - bt}}} \right)} \right)\]
\[ \Rightarrow v = \dfrac{k}{b}\dfrac{d}{{dt}}\left( {1 - {e^{ - bt}}} \right)\]
\[ \Rightarrow v = \dfrac{k}{b}\left( { - {e^{ - bt}}\left( { - b} \right)} \right)\]
\[ \Rightarrow v = k\,{e^{ - bt}}\]
Here, k and b are the constant quantities.
The above expression gives the speed of the particle at time t.
So, the correct answer is option (A).
Note: Be careful while differentiating the exponential terms. The derivative of the term
\[\dfrac{d}{{dx}}\left( {{e^{nx}}} \right) = n{e^{nx}}\] and not just \[{e^{nx}}\]. The expression
for the speed that we have derived only gives the speed of the particle at the instant of time
t and not the average velocity of the particle after time t. If you differentiate the expression
for the speed with respect to time t, you will get the acceleration of the particle at that
instant of time.
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