An automobile accelerates from rest at
1.3 m/s
2
for 19 s. The speed is then held
constant for 23 s, after which there is an acceleration of −0.9 m/s
2
until the automobile
stops.
What total distance was traveled?
Answer in units of km
I am still confused. For finalvelocity, where did you get 18?
while accelerating:
distance=1/2 1.3 19^2
finalvelocity=1.3*18
constant speed period.
distance= finalvelocityabove*23 seconds
deaccelerating period.
Vf^2=Vi^2 + 2ad where Vf=0 and Vi equals the finalvelocityabove. Solve for distance.
time deaccelerating: 0=Vi+at solve for t
add the total distances, and if you need, add the total times.
To find the total distance traveled by the automobile, we need to calculate the distances covered during each phase of motion and then sum them up.
Let's break down the problem into three phases:
Phase 1: Acceleration from rest
The initial velocity (u) is 0 m/s, the acceleration (a) is 1.3 m/s², and the time (t) is 19 s. We can use the equation of motion:
\(s = ut + \frac{1}{2}at²\)
Substituting the values, we get:
\(s_1 = 0(19) + \frac{1}{2}(1.3)(19)²\)
s₁ = 0 + 0.5 * 1.3 * 19²
s₁ = 0 + 0.5 * 1.3 * 361
s₁ = 0 + 0.5 * 469.3
s₁ = 0 + 234.65
s₁ = 234.65 m
Phase 2: Constant speed
The speed is held constant for 23 s, which means there is no acceleration. To find the distance covered during this phase, we use the formula:
\(s = ut\)
Since the speed is constant, the distance covered is given by:
\(s_2 = v \cdot t\)
Substituting the values, we get:
\(s_2 = 1.3 \times 23\)
s₂ = 1.3 * 23
s₂ = 29.9 m
Phase 3: Deceleration to a stop
The initial velocity is the same as the final velocity of the second phase (1.3 m/s), and the acceleration is given as -0.9 m/s². We can use the equation of motion again:
\(s = ut + \frac{1}{2}at²\)
The final velocity (v) is 0 m/s. Substituting the values, we get:
\(0 = 1.3(23 + t) + \frac{1}{2}(-0.9)(23 + t)²\)
Simplifying the equation would result in a quadratic equation, but instead of solving it using the quadratic formula, we can recognize that the time when the automobile comes to a stop is when the distance covered during Phase 3 (s₃) is equal to the distance covered during Phase 2 (s₂). So we can directly set:
\(s₃ = s₂\)
Now we can solve for 't':
\(1.3(23 + t) + \frac{1}{2}(-0.9)(23 + t)² = 29.9\)
Solving this equation will give us 't', which is the time it takes for the automobile to come to a stop.
Once we have 't', we can find the distance covered during Phase 3 using the equation:
\(s₃ = 1.3t + \frac{1}{2}(-0.9)t²\)
Finally, the total distance traveled is the sum of s₁, s₂, and s₃:
\(Total\ distance = s₁ + s₂ + s₃\)
Upon solving the equation and summing up the distances, we get the total distance traveled by the automobile in meters. To convert it to km, we divide by 1000:
\(Total\ distance\ in\ km = Total\ distance\ in\ m / 1000\)