Question

1. during a car race, a driver approaches a turn in the track traveling east. the driver hits the brakes for a total of 4 s, the car experiences an acceleration of -1.2 m/s² during this time period to reach a velocity of 26.8 m/s before entering the turn in the track. how fast was the car originally traveling down the track?

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v = v_0+at$, 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 $v_0$

$v_0=v - at$.

Step3: Substitute the given values

Given $v = 26.8\ m/s$, $a=- 1.2\ m/s^2$, and $t = 4\ s$. Then $v_0=26.8-(-1.2)\times4$.

Step4: Calculate the result

$v_0=26.8 + 4.8=31.6\ m/s$.

Answer:

$31.6\ m/s$