A car is moving in such a ways that the distance it covers , is given by the equations ` s = 4t^(2) +3t, ` where s, is in meters and t is in seconds . What would be the velocity and the accelerations of the car at time t=20 seconds?

Answer

To solve the problem, we need to find the velocity and acceleration of a car given its displacement equation \( s = 4t^2 + 3t \). We will follow these steps:

### Step 1: Find the velocity
The velocity \( v \) is the derivative of the displacement \( s \) with respect to time \( t \). We can express this mathematically as:

\[ v = \frac{ds}{dt} \]

Given the displacement equation:

\[ s = 4t^2 + 3t \]

We differentiate \( s \) with respect to \( t \):

\[ v = \frac{d}{dt}(4t^2 + 3t) \]

Using the power rule of differentiation:

\[ v = 8t + 3 \]

### Step 2: Calculate the velocity at \( t = 20 \) seconds
Now we substitute \( t = 20 \) seconds into the velocity equation:

\[ v = 8(20) + 3 \]

Calculating this gives:

\[ v = 160 + 3 = 163 \text{ m/s} \]

### Step 3: Find the acceleration
The acceleration \( a \) is the derivative of the velocity \( v \) with respect to time \( t \). We can express this mathematically as:

\[ a = \frac{dv}{dt} \]

From our previous result, we have:

\[ v = 8t + 3 \]

Now we differentiate \( v \) with respect to \( t \):

\[ a = \frac{d}{dt}(8t + 3) \]

Using the power rule of differentiation:

\[ a = 8 \]

### Step 4: Summary of results
At \( t = 20 \) seconds:
- The velocity of the car is \( 163 \text{ m/s} \).
- The acceleration of the car is \( 8 \text{ m/s}^2 \).

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