Video Transcript
A particle is initially at rest at the origin. It starts to move along the 𝑥-axis such that at time 𝑡 seconds its velocity is given by 𝑣 equal to 𝑡 multiplied by 27 minus 54𝑡 meters per second, where 𝑡 is greater than or equal to zero. What is the particle’s displacement when it comes to instantaneous rest?
In this question, we’re given an expression for the velocity of a particle in terms of time. We recall that we can convert expressions for displacement, velocity, and acceleration in terms of time by differentiating and integrating. We can differentiate an expression for displacement to obtain an expression for velocity. Likewise, differentiating an expression for velocity in terms of time gives us an expression for the acceleration. As integration is the inverse of differentiation, we can also use this to convert from one expression to another.
As already mentioned, in this question, we have an expression for the velocity. And we are trying to calculate its displacement when it comes to rest. We can find an expression for the displacement 𝑠 of 𝑡 by integrating 𝑣 of 𝑡 with respect to 𝑡. 𝑠 of 𝑡 is therefore equal to the integral of 𝑡 multiplied by 27 minus 54𝑡 with respect to 𝑡. Before integrating our expression, we will distribute the parentheses. This means that we need to integrate 27𝑡 minus 54𝑡 squared with respect to 𝑡. We can do this term by term. Integrating 27 𝑡 gives us 27𝑡 squared over two and integrating 54𝑡 squared gives us 54𝑡 cubed over three, which simplifies to 18𝑡 cubed. The displacement 𝑠 of 𝑡 is therefore equal to 27𝑡 squared over two minus 18 𝑡 cubed plus our constant of integration 𝐶.
We are told that the particle is initially at rest at the origin. This means that when 𝑡 is equal to zero, 𝑠 of 𝑡 equals zero. Substituting these values into our equation, we see that the constant of integration 𝐶 is also equal to zero. The displacement 𝑠 of 𝑡 is therefore equal to 27𝑡 squared over two minus 18𝑡 cubed. We want to calculate the value of 𝑠 of 𝑡 when the particle comes to instantaneous rest. This occurs when 𝑣 is equal to zero. Setting our expression for velocity equal to zero, we have 𝑡 multiplied by 27 minus 54𝑡 equals zero. This means that either 𝑡 equals zero or 27 minus 54𝑡 equals zero.
We can solve the second equation by adding 54𝑡 to both sides. We can then divide through by 54 such that 𝑡 is equal to 27 over 54, which simplifies to one-half. The particle comes to instantaneous rest when 𝑡 equals one-half. We can now substitute this value of 𝑡 into our expression for the displacement. 𝑠 of 𝑡 is equal to 27 multiplied by a half squared divided by two minus 18 multiplied by a half cubed. A half squared is equal to a quarter. And multiplying this by 27 and then dividing by two gives us 27 over eight. As a half cubed is one-eighth, the second term becomes 18 over eight. As the denominators are the same, we can then subtract the numerators, giving us nine over eight.
We can therefore conclude that the particle’s displacement when it comes to instantaneous rest is nine-eighths meters.
