an ambulance starts its motion from rest at a hospital and accelerates at a rate of 1.60 m/s^2 for 14.0 s. after that, it travels at constant speed for 70.0 s, and it the slows down at a rate of 3.5 m/s^2 until it stops (vf=o m/s) at the accident seen. find the total distance covered by the ambulance.
To find the total distance covered by the ambulance, we need to calculate the distance during each phase of its motion - the acceleration phase, the constant speed phase, and the deceleration phase.
1. Acceleration Phase:
Using the equation for distance during uniform acceleration:
d = (1/2) * a * t^2
where:
d = distance traveled
a = acceleration
t = time
Plugging in the values for the acceleration and time during the acceleration phase, we get:
d1 = (1/2) * 1.60 m/s^2 * (14.0 s)^2
2. Constant Speed Phase:
The distance covered during constant speed can be found using the formula:
d2 = v * t
where:
d2 = distance
v = velocity
t = time
Since the ambulance travels at constant speed, the velocity will remain constant throughout this phase. To find the velocity, we can use the equation for uniform acceleration:
v = a * t_initial
where:
v = velocity
a = acceleration
t_initial = initial time
Substituting the values for the initial time and acceleration during the acceleration phase, we have:
v = 1.60 m/s^2 * (14.0 s)
Now we can substitute this value for v and the time for the constant speed phase into the distance formula:
d2 = v * t = (1.60 m/s^2 * 14.0 s) * 70.0 s
3. Deceleration Phase:
During the deceleration phase, the goal is to slow down and eventually stop at the accident scene. We can use the formula for distance during uniform acceleration (deceleration in this case) again:
d3 = (1/2) * a * t^2
Using the values for acceleration and time for the deceleration phase, we have:
d3 = (1/2) * (-3.5 m/s^2) * (t_final)^2
Note that the acceleration during deceleration is negative because it opposes the motion.
To find the final time (t_final), we can use the formula:
v_final = v_initial + a * t
Since the ambulance stops at the accident scene, the final velocity (v_final) is zero. We can solve for t_final:
0 = v_initial + a * t_final
Substituting the values for v_initial and a, we get:
0 = 1.60 m/s^2 * (14.0 s) - 3.5 m/s^2 * t_final
Solving for t_final gives us:
14.0 s * 1.60 m/s^2 = 3.5 m/s^2 * t_final
t_final = (14.0 s * 1.60 m/s^2) / 3.5 m/s^2
Finally, we can substitute this value for t_final into the distance formula:
d3 = (1/2) * (-3.5 m/s^2) * [(14.0 s * 1.60 m/s^2) / 3.5 m/s^2]^2
Now we can calculate the total distance covered by adding the distances from each phase:
total distance = d1 + d2 + d3