Question

a race car moves such that its position fits the relationship x=(5m/s)t+(0.75m/s)t^3 where x measured in meters and t measured in second determine instantaneous velocity at t=4sec

Answer 1

To determine the instantaneous velocity of the race car at \( t = 4 \, \text{s} \), we need to find the derivative of the position function \( x(t) \) with respect to time \( t \).

The position function is given as:

\[
x(t) = (5 \, \text{m/s}) t + (0.75 \, \text{m/s}) t^3
\]

Now, we differentiate \( x(t) \) to find the velocity function \( v(t) \):

\[
v(t) = \frac{dx(t)}{dt} = \frac{d}{dt} \left( (5 \, \text{m/s}) t + (0.75 \, \text{m/s}) t^3 \right)
\]

Using the power rule, we differentiate each term:

1. The derivative of \( (5 \, \text{m/s}) t \) is \( 5 \, \text{m/s} \).
2. The derivative of \( (0.75 \, \text{m/s}) t^3 \) is \( 3 \cdot (0.75 \, \text{m/s}) t^2 = (2.25 \, \text{m/s}) t^2 \).

Combining these results:

\[
v(t) = 5 \, \text{m/s} + (2.25 \, \text{m/s}) t^2
\]

Next, we substitute \( t = 4 \, \text{s} \) into the velocity function to find the instantaneous velocity:

\[
v(4) = 5 \, \text{m/s} + (2.25 \, \text{m/s}) (4 \, \text{s})^2
\]

Calculating the \( t^2 \) term:

\[
(4 \, \text{s})^2 = 16 \, \text{s}^2
\]

Thus:

\[
v(4) = 5 \, \text{m/s} + (2.25 \, \text{m/s}) \cdot 16 \, \text{s}^2
\]
\[
v(4) = 5 \, \text{m/s} + 36 \, \text{m/s}
\]
\[
v(4) = 41 \, \text{m/s}
\]

Therefore, the instantaneous velocity at \( t = 4 \, \text{s} \) is \( \boxed{41 \, \text{m/s}} \).