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Question: A particle is moving in a straight line. Its displacement at time t is given by \(s\left( {{\text{in...

A particle is moving in a straight line. Its displacement at time t is given by s(in m)=−4t2+2ts\left( {{\text{in m}}} \right) = - 4{t^2} + 2t ,then its velocity and acceleration at time t=12t = \dfrac{1}{2} second are?

Explanation

Solution

To solve this type of question, one must know basics of differentiation, we will simply apply differentiation as we know that rate of change of displacement is velocity and we will simply put the obtained value of time in the equation to obtain the velocity and for acceleration similarly we will differentiate the obtained equation of velocity to get the required result.

Complete step by step answer:
According to the question the given equation is,
s(in m)=−4t2+2ts\left( {{\text{in m}}} \right) = - 4{t^2} + 2t
And we know that the derivative of displacement is displacement.So,
v(t)=dsdt=−4t2+2tv(t) = \dfrac{{ds}}{{dt}} = - 4{t^2} + 2t
⇒v(t)=−8t+2\Rightarrow v(t) = - 8t + 2 -----(1)
And we have to find the velocity at time t=12t = \dfrac{1}{2}

So, now putting the values of tt in above equation,
v(t)=−8t+2 ⇒v(12)=−8×12+2 ⇒v(12)=−2 v(t) = - 8t + 2 \\\ \Rightarrow v\left( {\dfrac{1}{2}} \right) = - 8 \times \dfrac{1}{2} + 2 \\\ \Rightarrow v\left( {\dfrac{1}{2}} \right) = - 2 \\\
Hence, the velocity at the given moment of time is −2m s−1 - 2m{\text{ }}{s^{ - 1}}. For acceleration, we know that the rate of change of velocity is acceleration or in the derivative we can write,
a(t)=dvdta\left( t \right) = \dfrac{{dv}}{{dt}}
So, in the above solution we have equation 1 as the equation of velocity.Now, differentiating,
a(t)=dvdt=−8a(t) = \dfrac{{dv}}{{dt}} = - 8
So, we can conclude from the above result that acceleration is constant throughout the motion.

Hence acceleration of the body is −8m s−2 - 8m{\text{ }}{s^{ - 2}}.

Note: Note that the velocity is the rate of change of displacement whereas the acceleration is the rate of change of velocity. Both the physical quantity is vector quantity because they both have magnitude as well as direction. Velocity may or may not be zero and in case of acceleration, as it is defined as a change in velocity or speed, acceleration cannot be zero. This explains why there should be some movement in order to accelerate.