The distance travelled by a bus in t seconds after the brakes are applied is $1 + 2t - 2{t^2}$ metres. The distance travelled by the bus before it stops is equal to:
A. $0.5m$
B. $1m$
C. $1.5m$
D. $2.5m$
Answer
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Hint: In the question, we are provided with the expression for distance travelled by the bus after brakes are applied. So, we first find the instant when the speed of the bus becomes zero. We find the expression for bus speed by differentiating the expression for distance travelled by bus. Then, we substitute the value of time in the expression for distance to find the distance travelled by bus before it comes to rest.
Complete step by step answer:
In the problem, we have the expression $s = 1 + 2t - 2{t^2}$. So, we differentiate both sides of the expression. We should know the power rule of differentiation, we get,
$ \Rightarrow \dfrac{{ds}}{{dt}} = 2 - 4t$
Now, we know that the derivative of distance with respect to time gives us speed. So, to find the time when the speed of the bus becomes zero. So, equating $\dfrac{{ds}}{{dt}}$ to zero.
$ \Rightarrow \dfrac{{ds}}{{dt}} = 2 - 4t = 0$
So, shifting the term consisting $t$ to the right side of equation, we get,
$ \Rightarrow 2 = 4t$
$ \Rightarrow t = \dfrac{1}{2}$
So, we have to find the distance travelled by the bus in $\dfrac{1}{2}$ second till it stops. So, substituting the value of time in the expression of distance travelled by bus, we get,
$ \Rightarrow 1 + 2\left( {\dfrac{1}{2}} \right) - 2{\left( {\dfrac{1}{2}} \right)^2}$
Evaluating the powers of $\left( {\dfrac{1}{2}} \right)$, we get,
\[ \Rightarrow 1 + 1 - \left( {\dfrac{1}{2}} \right)\]
\[ \Rightarrow 2 - \dfrac{1}{2}\]
\[ \therefore \dfrac{3}{2} = 1.5m\]
So, the distance travelled by the bus before it stops is equal to \[1.5\,m\].
Hence, option (C) is the correct answer.
Note: We must know that the derivative of distance travelled by a body gives the instantaneous speed of the body. We must know the rules of differentiation to solve the problem. Care should be taken while doing the calculations to be sure of the answer. One must know that the derivative of a constant is zero.
Complete step by step answer:
In the problem, we have the expression $s = 1 + 2t - 2{t^2}$. So, we differentiate both sides of the expression. We should know the power rule of differentiation, we get,
$ \Rightarrow \dfrac{{ds}}{{dt}} = 2 - 4t$
Now, we know that the derivative of distance with respect to time gives us speed. So, to find the time when the speed of the bus becomes zero. So, equating $\dfrac{{ds}}{{dt}}$ to zero.
$ \Rightarrow \dfrac{{ds}}{{dt}} = 2 - 4t = 0$
So, shifting the term consisting $t$ to the right side of equation, we get,
$ \Rightarrow 2 = 4t$
$ \Rightarrow t = \dfrac{1}{2}$
So, we have to find the distance travelled by the bus in $\dfrac{1}{2}$ second till it stops. So, substituting the value of time in the expression of distance travelled by bus, we get,
$ \Rightarrow 1 + 2\left( {\dfrac{1}{2}} \right) - 2{\left( {\dfrac{1}{2}} \right)^2}$
Evaluating the powers of $\left( {\dfrac{1}{2}} \right)$, we get,
\[ \Rightarrow 1 + 1 - \left( {\dfrac{1}{2}} \right)\]
\[ \Rightarrow 2 - \dfrac{1}{2}\]
\[ \therefore \dfrac{3}{2} = 1.5m\]
So, the distance travelled by the bus before it stops is equal to \[1.5\,m\].
Hence, option (C) is the correct answer.
Note: We must know that the derivative of distance travelled by a body gives the instantaneous speed of the body. We must know the rules of differentiation to solve the problem. Care should be taken while doing the calculations to be sure of the answer. One must know that the derivative of a constant is zero.
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