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9 a driver of a car traveling at 35.0 m/s applies the brakes causing a …

Question

9 a driver of a car traveling at 35.0 m/s applies the brakes causing a uniform acceleration of -3.0 m/s². how long does it take the car to accelerate to a final speed of 10.0 m/s? record your answer and fill in the bubbles on your answer document. be sure to use the correct place value. 10 an automobile with an initial speed of 4.3 m/s accelerates uniformly at a rate of 3.0 m/s². find the final speed and the displacement after 5.0 seconds. record your answer and fill in the bubbles on your answer document. be sure to use the correct place value. 11 an automobile with an initial speed of 4.3 m/s accelerates uniformly at a rate of 3.0 m/s². find the final speed and the displacement after 5.0 seconds. record your answer and fill in the bubbles on your answer document. be sure to use the correct place value.

Explanation:

Step1: Identify the kinematic - equation

We use the equation \(v = v_0+at\) for question 9, where \(v\) is the final velocity, \(v_0\) is the initial velocity, \(a\) is the acceleration and \(t\) is the time.

Step2: Rearrange the equation for time

From \(v = v_0 + at\), we can solve for \(t\): \(t=\frac{v - v_0}{a}\). Given \(v_0 = 35.0\ m/s\), \(v = 10.0\ m/s\) and \(a=- 3.0\ m/s^2\).

Step3: Calculate the time

Substitute the values into the formula: \(t=\frac{10.0 - 35.0}{-3.0}=\frac{- 25.0}{-3.0}\approx8.33\ s\).

For questions 10 and 11:

Step1: Use the velocity - time equation for final speed

The equation \(v = v_0+at\). For \(v_0 = 4.3\ m/s\), \(a = 3.0\ m/s^2\) and \(t = 5.0\ s\), then \(v=4.3+3.0\times5.0\).

Step2: Calculate the final speed

\(v=4.3 + 15.0=19.3\ m/s\).

Step3: Use the displacement - time equation

The displacement equation is \(x=v_0t+\frac{1}{2}at^{2}\). Substitute \(v_0 = 4.3\ m/s\), \(a = 3.0\ m/s^2\) and \(t = 5.0\ s\) into it: \(x=4.3\times5.0+\frac{1}{2}\times3.0\times5.0^{2}\).

Step4: Calculate the displacement

First, calculate \(4.3\times5.0 = 21.5\) and \(\frac{1}{2}\times3.0\times5.0^{2}=\frac{3}{2}\times25 = 37.5\). Then \(x=21.5 + 37.5=59.0\ m\).

Answer:

Question 9: \(8.33\ s\)
Question 10: Final speed: \(19.3\ m/s\), Displacement: \(59.0\ m\)
Question 11: Final speed: \(19.3\ m/s\), Displacement: \(59.0\ m\)