A car initially at rest travels for 2 minutes with a uniform acceleration of 0.3 m/s^2,after which the speed is kept constant until the car is brought to rest with a uniform retardation of 0.6 m/s^2.If the total distance covered is 4500m,what is the time taken for the journey?
First leg of trip:
a = .3
v = .3t + c
when t = 0 , v = 0, so c = 0
v = .3t
s = .15t^2 + k
when t = 0, s = 0, so k = 0
so s = .15t^2 is the distance after t seconds
when t = 2 min = 120sec
v = .3(120) = 36 m/s
s = .15(120^2) m = 2160 m
so the car went 2160 m before hitting the brakes, and had a velocity of
36 m/s
Middle part of trip:
constant speed of 36 m/s for t seconds
distance covered = 36t
3rd leg of trip
a = -.6
v = -.6t + c
so initially t = 0 again, v = .3(120) = 36 m/s <---- velocity from first trip
36 = c
v = -.6t + 36
s = - .3t^2 + 36t + k
when t = 0, distance covered by 2nd trip was 0
k = 0
s = -.3t^2 + 36t
we want the car to stop, so v = 0
-.6t + 36 = 0
t = 60
so it takes 60 seconds to stop, and the distance covered in that time
= -.3(60^2) + 36(60)
= -1080 + 2160 = 1080 m
total distance = 2160 + 36t + 1080 = 4500
36t = 4500-2160-1080 = 1260
t = 35 seconds
total time = 120+35+60 seconds = 215 seconds or 3 minutes and 35 seconds
To find the time taken for the journey, we need to divide the problem into two parts: the first part where the car accelerates, and the second part where it decelerates.
Let's start by finding the time taken during the acceleration phase.
First, let's determine the final velocity of the car during the acceleration phase. We can use the equation:
v = u + at
where:
v is the final velocity
u is the initial velocity (0 m/s since the car is initially at rest)
a is the acceleration (0.3 m/s^2)
t is the time taken during the acceleration phase
Since the car starts from rest, its initial velocity (u) is 0 m/s. Therefore, the equation simplifies to:
v = 0 + (0.3)(t)
v = 0.3t
Now, let's determine the distance covered during the acceleration phase. We can use the equation:
s = ut + (1/2)at^2
where:
s is the distance covered
u is the initial velocity (0 m/s)
a is the acceleration (0.3 m/s^2)
t is the time taken during the acceleration phase
Since the car starts from rest, its initial velocity (u) is 0 m/s. Therefore, the equation simplifies to:
s = (1/2)(0.3)(t^2)
s = 0.15t^2
Now, let's move on to the deceleration phase.
During the deceleration phase, the car's acceleration is negative (-0.6 m/s^2) because it is decelerating. We can use the same equations, but with the negative acceleration value.
The final velocity during the deceleration phase is given by:
v = u + at
Since the car's final velocity is 0 at the end of the journey, the equation becomes:
0 = v - (0.6)(t)
0 = v - 0.6t
Let's solve for v in terms of t:
v = 0.6t
Now, let's determine the distance covered during the deceleration phase. Using the same equation as before:
s = ut + (1/2)at^2
Since the initial velocity (u) is the final velocity during the acceleration phase (0.3t), the equation becomes:
s = (0.3t)t + (1/2)(-0.6)(t^2)
s = 0.3t^2 - 0.3t^2
s = 0
Since the distance covered during the deceleration phase is 0, we know that all of the distance (4500 m) was covered during the acceleration phase.
Therefore, the equation becomes:
4500 = 0.15t^2
Rearranging the equation, we get:
t^2 = 4500 / 0.15
t^2 = 30000
Taking the square root of both sides, we find:
t = √30000
t ≈ 173.21 seconds
Therefore, the time taken for the entire journey is approximately 173.21 seconds.
Write any two different between mks and CGS system
Well, well, well! Looks like this car had quite the adventure! Let's break it down, shall we?
First, let's find the distance covered during the acceleration phase. We can use the formula: distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Since the car starts from rest, the initial velocity is 0. The acceleration is 0.3 m/s^2, and the time is 2 minutes (or 120 seconds). Plugging in these values, we get:
distance1 = 0.5 * 0.3 * (120^2)
Now, let's find the distance covered during the deceleration phase. Again, using the same formula but with a negative acceleration this time (since it's deceleration). The acceleration is -0.6 m/s^2, and we don't know the time, so let's call it "t":
distance2 = (0 * t) + (0.5 * -0.6 * t^2)
The total distance covered is given as 4500m, so we can write:
distance1 + distance2 = 4500
Substituting the values we found earlier:
0.5 * 0.3 * (120^2) + (0.5 * -0.6 * t^2) = 4500
Solving this equation will give us the time taken for the journey. But hey, chances are my answer might be a bit too anticlimactic, so I'll leave the arithmetic to you! Go on, crunch those numbers, you got this!
To find the time taken for the journey, we need to break down the motion into three stages: acceleration, constant speed, and deceleration.
Stage 1: Acceleration
Since the car starts from rest, we can use the following equation of motion to find the distance covered during acceleration:
\[s_1 = u \cdot t + \frac{1}{2} \cdot a \cdot t^2\]
where:
- \(s_1\) is the distance covered during acceleration
- \(u\) is the initial velocity (0 m/s)
- \(t\) is the time taken during acceleration (unknown)
- \(a\) is the acceleration (0.3 m/s^2)
Rearranging the equation, we get:
\[s_1 = \frac{1}{2} \cdot a \cdot t^2\]
Stage 2: Constant Speed
During this stage, the car is moving with a constant speed. Since the speed doesn't change, the distance covered can be calculated using:
\[s_2 = \text{constant speed} \cdot \text{time}\]
Stage 3: Deceleration
Finally, we need to find the time taken during deceleration. Similar to Stage 1, we can use the equation of motion:
\[s_3 = u \cdot t + \frac{1}{2} \cdot a \cdot t^2\]
where:
- \(s_3\) is the distance covered during deceleration
- \(u\) is the initial velocity during deceleration (unknown)
- \(t\) is the time taken during deceleration (unknown)
- \(a\) is the acceleration during deceleration (-0.6 m/s^2)
Rearranging the equation, we get:
\[s_3 = u \cdot t - \frac{1}{2} \cdot a \cdot t^2\]
Applying the given information:
- Total distance covered (s) = 4500 m
- Acceleration (a1) = 0.3 m/s^2
- Deceleration (a2) = -0.6 m/s^2
After summing up the distances covered in each stage, we have:
\[s_1 + s_2 + s_3 = 4500\]
Substituting the equations for each stage:
\[\frac{1}{2} \cdot a_1 \cdot t^2 + \text{constant speed} \cdot \text{time} + u \cdot t - \frac{1}{2} \cdot a_2 \cdot t^2 = 4500\]
Simplifying the equation, we get:
\[\left(\frac{1}{2} \cdot a_1 - \frac{1}{2} \cdot a_2\right) \cdot t^2 + (u + \text{constant speed}) \cdot t - 4500 = 0\]
Now, we can solve this quadratic equation for time (t).