a car traveling at a velocity of 90 km/h, when you see a dog step into the road 50 m ahead. you hit the breaks hard to get maximum deceleration of 7.5 m/s^2. how far will you travel before stopping?
Vo = 90000m/h * (1/3600)h/s = 25m/s.
Vf^2 = Vo^2 + 2ad,
d = (Vf^2 - Vo^2) / 2a,
d = (0 - (25)^2) / -15 = 41.7m. before stopping.
To calculate the distance the car will travel before stopping, we can use the equations of motion.
The first step is to convert the initial velocity from km/h to m/s.
1 km/h is equal to 0.28 m/s (approximately).
So, the initial velocity of 90 km/h can be converted to:
90 km/h * 0.28 m/s = 25.2 m/s (approximately).
To find the stopping distance, we can use the formula:
\( a = \frac{{v_f - v_i}}{{t}} \),
where:
a is the acceleration (deceleration in this case),
v_f is the final velocity (which is 0, as the car stops),
v_i is the initial velocity (25.2 m/s), and
t is the time taken to come to a stop.
To find the time taken to stop, we use the formula:
\( t = \frac{{v_f - v_i}}{{a}} \).
Substituting the values, we get:
\( t = \frac{{0 - 25.2}}{{-7.5}} \).
Note that we use a negative value for acceleration (-7.5 m/s^2) as it is a deceleration.
Now, solving for t:
\( t = \frac{{-25.2}}{{-7.5}} = 3.36 \) seconds (approximately).
Now, to find the distance traveled, we use the formula:
\( d = v_i t + \frac{1}{2} a t^2 \).
Substituting the values we have:
\( d = 25.2 * 3.36 + \frac{1}{2} * (-7.5) * (3.36^2) \).
Simplifying the equation:
\( d = 84.672 + \frac{1}{2} * (-7.5) * 11.2896 \).
Evaluating the expression:
\( d = 84.672 + (-42.432) \).
Finally, we get the distance traveled:
\( d = 42.24 \) meters (approximately).
Therefore, the car will travel approximately 42.24 meters before stopping.