Question
A particle moves in a straight line such that its velocity, $vms^{-1}$ at time t seconds is given by $v (t) = 1 + e^{-t} -e^{-sin(2t)}$ for $0\leq t\leq2$
(a) Find the velocity of the particle at $t = 2$.
(b) Find the maximum velocity of the particle.
(c) Find the acceleration of the particle at the instant it changes direction.
▶️Answer/Explanation
Detail Solution
part(a)
Step 1: Substitute \(t = 2\) into the velocity function.
The velocity function is given by \(v(t) = 1 + e^{-t} – e^{-\sin(2t)}\).
Substituting \(t = 2\):
\(v(2) = 1 + e^{-2} – e^{-\sin(4)}\).
Step 2: Calculate \(e^{-2}\) and \(e^{-\sin(4)}\).
Using a calculator, we find:
\(e^{-2} \approx 0.1353\) and \(\sin(4) \approx -0.7568\), thus \(e^{-\sin(4)} \approx e^{0.7568} \approx 2.126\).
Step 3: Combine the values to find \(v(2)\).
\(v(2) = 1 + 0.1353 – 2.126 \approx 1 + 0.1353 – 2.126 \approx -0.9907\).
The velocity at \(t = 2\) is approximately \(-0.9907\).
The answer is -0.9907
part(b)
From the graph of velocity function $v (t) = 1 + e^{-t} -e^{-sin(2t)}$
Step 1: Find the maximum velocity by analyzing the velocity function.
To find the maximum velocity, we need to find the critical points by taking the derivative of \(v(t)\) and setting it to zero.
The derivative is given by \(v'(t) = -e^{-t} + 2e^{-\sin(2t)}\cos(2t)\).
Step 2: Set the derivative equal to zero.
Setting \(v'(t) = 0\):
\(-e^{-t} + 2e^{-\sin(2t)}\cos(2t) = 0\).
This equation is complex to solve analytically, so we will evaluate \(v(t)\) at the endpoints \(t = 0\) and \(t = 2\) and check for critical points in the interval \(0 \leq t \leq 2\).
Step 3: Calculate \(v(0)\) and \(v(2)\).
At \(t = 0\):
\(v(0) = 1 + e^{0} – e^{0} = 1\).
At \(t = 2\):
As calculated previously, \(v(2) \approx -0.9907\).
Step 4: Evaluate \(v(t)\) at critical points numerically.
Using numerical methods or graphing tools,
After analyzing the graph , at t=0.40 occurs maximum value for the interval $ 0\leq t \leq 2$
we find that the maximum velocity occurs at \(t \approx 0.40\) where \(v(0.40) \approx 1.18\).
Thus, the maximum velocity is approximately \(1.18\).
The answer is: 1.18
part(c)
Step 1: Determine when the particle changes direction.
The particle changes direction when the velocity \(v(t) = 0\).
Setting the velocity function to zero:
\(1 + e^{-t} – e^{-\sin(2t)} = 0\).
Step 2: Solve for \(t\) numerically.
Using numerical methods, we find that \(v(t) = 0\) occurs at approximately \(t \approx 1.65\).
Step 3: Find the acceleration at this time.
The acceleration \(a(t)\) is the derivative of the velocity function:
\(a(t) = v'(t) = -e^{-t} + 2e^{-\sin(2t)}\cos(2t)\).
Substituting \(t \approx 1.65\):
Calculate \(a(1.65)\).
Using numerical values, we find \(a(1.65) \approx -2.5\).
The answer is -2.50
————Markscheme—————–
(a) $v = -0.996114…$
$v = -0.996 \text{ (ms}^{-1}\text{)}$
(b) considers $v'(t) = 0$
$t = 0.405833…$
$v_{max} = 1.18230…$
$v_{max} = 1.18 \text{ (ms}^{-1}\text{)}$
(c) recognises that the particle changes direction when $v = 0$
finds acceleration for their value of $t$ for which $v(t) = 0$
$v'(1.65840…)$
$a = -2.53487…$
$a = -2.53 \text{ (ms}^{-2}\text{)}$