Final answer:
The car was initially moving at 15.6 m/s and decelerated at -2.5 m/s² for 12.3 seconds, which mathematically leads to a final speed of -15.15 m/s. This result signifies that the car stops before 12.3 seconds and continues in the opposite direction. Option E.
Step-by-step explanation:
To find out how fast the car is going after slowing down, we can use the equation of motion that relates initial velocity, acceleration, and time.
The equation is:
v = u + at
Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
Given that the initial velocity u is 15.6 m/s, acceleration a is -2.5 m/s² (since it's deceleration), and time t is 12.3 seconds, we can plug these values into the equation:
v = 15.6 m/s + (-2.5 m/s²) × 12.3 s
Calculation:
v = 15.6 m/s - (30.75 m/s)
v = -15.15 m/s
Since the velocity cannot be negative when considering only its magnitude
This implies that the car has stopped and started moving in the opposite direction (or it's an indication of the direction of motion compared to the original direction).
However, the magnitude of the new velocity is 15.15 m/s.
But because the question pertains to how fast the car is going before it has fully stopped (without considering a change of direction),
The car comes to a stop before 12.3 seconds, as the negative result suggests it has decelerated to zero and then continues to 'increase' velocity in the opposite direction.
Hence, the right answer is option E.
The complete question is:
A car traveling at 15.6 m/s hits the brakes to slow down before turning into a neighborhood. Determine how fast it is going after slowing down at a rate of -2.5 m/s^2 for 12.3 seconds. (round to the nearest tenth).
a. 6.2 m/s
b. 8.7 m/s
c. 12.0 m/s
d. 18.5 m/s
e) -15.15 m/s