Question

A car starts from rest with acceleration $a=6t (ms^{-2})$. Its velocity and displacement after 4 seconds are ________.

• A. 48, 64
• B. 16, 24
• C. 16, 38
• D. 24, 32
• E. 32, 24

Hint:

$v = \int a dt$ and $s = \int v dt$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
We are given a time-varying acceleration and need to find the velocity and displacement. This requires integration. Velocity is the integral of acceleration with respect to time, and displacement is the integral of velocity with respect to time. We must use the initial conditions (starting from rest) to find the constants of integration.
Step 2: Key Formula or Approach
1. Velocity (\(v\)): Since \(a = \frac{dv}{dt}\), we can find the velocity by integrating the acceleration: \(v(t) = \int a(t) dt\).
2. Displacement (\(s\)): Since \(v = \frac{ds}{dt}\), we can find the displacement by integrating the velocity: \(s(t) = \int v(t) dt\).
Initial conditions: The car starts from rest, which means at \(t=0\), the initial velocity \(v(0) = 0\). We also assume the initial displacement \(s(0) = 0\).
Step 3: Detailed Explanation
1. Find the velocity function v(t).
We are given \(a(t) = 6t\).
\[ v(t) = \int a(t) dt = \int 6t \, dt \] Using the power rule for integration:
\[ v(t) = 6 \frac{t^2}{2} + C_1 = 3t^2 + C_1 \] where \(C_1\) is the constant of integration.
To find \(C_1\), we use the initial condition \(v(0) = 0\).
\[ v(0) = 3(0)^2 + C_1 = 0 \implies C_1 = 0 \] So, the velocity function is \(v(t) = 3t^2\).
2. Calculate the velocity at t = 4 seconds.
\[ v(4) = 3(4)^2 = 3(16) = 48 \, \text{ms}^{-1} \] 3. Find the displacement function s(t).
Now we integrate the velocity function to find displacement.
\[ s(t) = \int v(t) dt = \int 3t^2 \, dt \] Using the power rule for integration:
\[ s(t) = 3 \frac{t^3}{3} + C_2 = t^3 + C_2 \] where \(C_2\) is the constant of integration.
We assume the initial displacement is zero, \(s(0) = 0\).
\[ s(0) = (0)^3 + C_2 = 0 \implies C_2 = 0 \] So, the displacement function is \(s(t) = t^3\).
4. Calculate the displacement at t = 4 seconds.
\[ s(4) = (4)^3 = 64 \, \text{m} \] Step 4: Final Answer
After 4 seconds, the velocity is 48 ms\(^{-1}\) and the displacement is 64 m.