Question

while traveling south on an expressway, a car traveling 60 mph (miles per hour) slows down to 30 mph in 12 minutes due to traffic conditions. calculate the acceleration. a = -150 mi/h² dropdown options: w, n/a, n, e, s

Explanation:

Step1: Recall the acceleration formula

Acceleration \( a \) is given by the change in velocity \( \Delta v \) divided by the change in time \( \Delta t \), i.e., \( a=\frac{\Delta v}{\Delta t} \).

Step2: Determine initial and final velocities

Initial velocity \( v_i = 60 \) mph, final velocity \( v_f = 30 \) mph. So, \( \Delta v=v_f - v_i=30 - 60=- 30 \) mph.

Step3: Convert time to hours

Time \( \Delta t = 12 \) minutes. Since 1 hour = 60 minutes, \( \Delta t=\frac{12}{60}=0.2 \) hours.

Step4: Calculate acceleration

Substitute \( \Delta v=-30 \) mph and \( \Delta t = 0.2 \) h into the acceleration formula: \( a=\frac{-30}{0.2}=- 150 \) \( \text{mi/h}^2 \). The direction: since the car is moving south and decelerating (slowing down), the acceleration is opposite to the direction of motion? Wait, no, the question's dropdown for direction: the car is moving south, and it's slowing down, so acceleration is north? Wait, no, let's think again. Wait, when a car is moving south and slows down, the acceleration is in the north direction (opposite to velocity direction). But wait, the problem's dropdown has S, N, etc. Wait, but the acceleration magnitude is - 150 \( \text{mi/h}^2 \), and the direction: since velocity is south (let's take south as positive), then \( v_i = 60 \) (south), \( v_f = 30 \) (south), so \( \Delta v=30 - 60=-30 \), so acceleration is \( \frac{-30}{0.2}=-150 \), which would mean acceleration is north (negative if south is positive). But wait, maybe the direction here: the car is moving south, and deceleration (negative acceleration) would be north? Wait, but let's check the time conversion again. Wait, 12 minutes is \( \frac{12}{60}=0.2 \) hours. Then \( \Delta v = 30 - 60=-30 \) mph. \( a=\frac{-30}{0.2}=-150 \) \( \text{mi/h}^2 \). The direction: since the car is moving south, and acceleration is negative (if we took south as positive), so the direction of acceleration is north? But the dropdown has N as an option. Wait, but maybe the problem's coordinate system: if we consider south as positive, then acceleration is negative, so direction is north. But let's confirm the calculation first. The acceleration magnitude is 150 \( \text{mi/h}^2 \) (deceleration), and direction is north (opposite to motion direction when slowing down).

Answer:

The acceleration is \(-150\) \( \text{mi/h}^2 \) with direction \( \text{N} \) (North). So the filled values are \( a = - 150\) \( \text{mi/h}^2 \) and direction \( \text{N} \).