We Crawled along through the city traffic at 6m/s for five minutes,then we left the city ,&we gradually accelerated to 24m/s in 20s. we kept going at this speed for 2minutes ,but then i noticed an accident on the road ahead and i braked , so that we came to ahalt in 8s. then find the car's acceleration & deceleration ??? & draw the graph to represent this journey???
draw the graph
draw graph
first 300 seconds (five min)
acceleration = 0
velocity = 6 m/s
x = 0 + 6 t linear
v = 6
at 300 seconds x = 6*300 = 1,800
second 20 seconds, t = 300 to t = 320
initial x = 1,800
initial speed = 6
final speed = 24
acceleration = (24-6)/20 = 18/20 = 0.9 m/s^2
final x = 1800 + (24+6)/2 * 20 = 1800 + 300 = 2,100
third segment
initial x = 2,100 m
speed = 24 m/s constant no acceleration
for 120 seconds so final time = 320+120 = 440 s
final x = 2,100 + 120*24 = 2100+2880 = 4980
fourth segment
initial x = 4980
initial v = 24
final v = 0
average v = 12 m/s
x = 4980 + 12 * 8 = 4980 + 96 = 5076
acceleration = -24/8 = -3 m/s^2
Wesd
Well, it seems like you had quite an eventful journey! Let's break it down and find the car's acceleration and deceleration.
First, let's convert the time into seconds for easier calculations:
5 minutes = 5 × 60 = 300 seconds
2 minutes = 2 × 60 = 120 seconds
8 seconds = 8 seconds
Acceleration:
The initial speed is 6 m/s, and it takes 300 seconds to reach a speed of 24 m/s.
Using the formula for acceleration (a = Δv/Δt), we have:
a = (24 - 6) / 300
a = 18 / 300
a = 0.06 m/s^2
Deceleration:
To find the deceleration, we need to calculate the change in velocity (Δv) and divide it by the time interval (Δt).
The final velocity is 0 m/s since the car comes to a halt.
Using the formula for deceleration, we have:
deceleration = Δv / Δt
deceleration = (24 - 0) / 8
deceleration = 24 / 8
deceleration = 3 m/s^2
As for the graph, I'm just a text bot, so I can't draw images. However, I can describe it for you!
Start the graph at (0, 0) as the initial position of the car.
Draw a straight line segment with a slope of 0.06 m/s^2, ascending from (0, 0) to (300, 24) on the x-axis.
Then, draw a horizontal line segment at a height of 24 m/s, extending from (300, 24) to (420, 24).
Finally, draw a straight line segment with a slope of -3 m/s^2, descending from (420, 24) to (428, 0) on the x-axis.
This should give you a rough graphical representation of your journey. Just don't forget to label your axes and units!
To find the car's acceleration and deceleration, we need to use the formulas of motion. Let's break down the given information step by step:
Step 1: Calculate the acceleration during the first phase of travel.
Given:
Initial velocity (u) = 6 m/s
Time (t) = 5 minutes = 300 seconds
Final velocity (v) = ?
We can use the formula: v = u + at, where a is the acceleration.
Since we have the initial velocity (u) and time (t), and the final velocity (v) is unknown, we need to rearrange the formula as follows:
a = (v - u)/t
Using the values: a = (v - 6 m/s)/(300 s)
Step 2: Calculate the acceleration during the second phase.
Given:
Initial velocity (u) = 6 m/s
Time (t) = 20 seconds
Final velocity (v) = 24 m/s
Acceleration (a) = ?
We can again use the formula: v = u + at, to find the acceleration.
Rearranging the formula, we have:
a = (v - u)/t
Using the given values: a = (24 m/s - 6 m/s)/(20 s)
Step 3: Calculate the deceleration.
Given:
Time (t) = 8 seconds
Initial velocity (u) = 24 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = ?
Using the same formula, a = (v - u)/t, we can substitute the given values:
a = (0 m/s - 24 m/s)/(8 s)
Now that we have the values for the accelerations and deceleration, we can proceed to plot the graph representing this journey.