Question

from rest, we step on the gas of our ferrari, providing a force f for 4 secs, speeding it up to a final speed v. if the applied force were only 1/2 f, how long would it have to be applied to reach the same final speed?

Explanation:

Step1: Use the impulse - momentum theorem

The impulse - momentum theorem states that \(J = \Delta p\). Initially, \(p_{i}=0\) (starts from rest). When force \(F\) is applied for time \(t_{1} = 4\space s\), the impulse \(J_{1}=F\times t_{1}\) and the final momentum \(p_{f}=mv\). So, \(F\times4=mv\).

Step2: Analyze the second - force case

Let the time for the force \(F_{2}=\frac{1}{2}F\) be \(t_{2}\). The impulse \(J_{2}=\frac{1}{2}F\times t_{2}\), and since the final momentum \(p_{f} = mv\) (same final speed, so same momentum change), we have \(\frac{1}{2}F\times t_{2}=mv\).

Step3: Equate the two expressions for \(mv\)

From \(F\times4 = mv\) and \(\frac{1}{2}F\times t_{2}=mv\), we can substitute \(mv = 4F\) into the second equation: \(\frac{1}{2}F\times t_{2}=4F\). Divide both sides by \(F\) (assuming \(F
eq0\)), we get \(\frac{1}{2}t_{2}=4\).

Step4: Solve for \(t_{2}\)

Multiply both sides of \(\frac{1}{2}t_{2}=4\) by \(2\). So, \(t_{2}=8\space s\).

Answer:

\(8\space s\)