Question

practice time!

1. a car accelerates from a standstill to 32 m/s in 11 s. what is its acceleration?
2. a car accelerates from 25 mi/hr to 55 mi/hr in 0.033 hr. what is its acceleration?
3. a runner has a velocity of 11.1 m/s for 9.0 seconds until he finishes the race and comes to a stop. what is his acceleration?
4. a motorist chist has an acceleration of 5.6 mi/hr². if his initial velocity is 45 mi/hr, what is his final velocity after 4.0 hours?

Explanation:

Step1: Recall the acceleration formula

The formula for acceleration \( a \) is \( a=\frac{v_f - v_i}{t} \), where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( t \) is the time.

Step2: Identify values for the first problem

For the first problem, \( v_i = 0\space m/s \) (standstill), \( v_f = 32\space m/s \), and \( t = 11\space s \).

Step3: Substitute values into the formula

Substitute into \( a=\frac{v_f - v_i}{t} \): \( a=\frac{32 - 0}{11}=\frac{32}{11}\approx2.91\space m/s^2 \).

Step4: Solve the second problem

For the second problem, \( v_i = 25\space mi/hr \), \( v_f = 55\space mi/hr \), \( t = 0.033\space hr \).
\( a=\frac{55 - 25}{0.033}=\frac{30}{0.033}\approx909.09\space mi/hr^2 \).

Step5: Solve the third problem

For the third problem, \( v_i = 11.1\space m/s \), \( v_f = 0\space m/s \) (stops), \( t = 9.0\space s \).
\( a=\frac{0 - 11.1}{9.0}=\frac{- 11.1}{9.0}\approx - 1.23\space m/s^2 \) (negative for deceleration).

Step6: Solve the fourth problem (assuming it's \( a = 5.6\space mi/hr^2 \), \( v_i = 45\space mi/hr \), \( t = 4.0\space hr \))

Use \( v_f=v_i + at \).
\( v_f = 45+(5.6\times4.0)=45 + 22.4 = 67.4\space mi/hr \).

Answer:

1. Acceleration: \(\approx 2.91\space m/s^2\)
2. Acceleration: \(\approx 909.09\space mi/hr^2\)
3. Acceleration: \(\approx - 1.23\space m/s^2\)
4. Final velocity: \( 67.4\space mi/hr \) (assuming the fourth problem's details as inferred)