A car weighing 1200kg was moving at a speed of 15m/s.breaks were applied and it stopped 6s latter.draw a graph showing the results
To draw a graph showing the results of the car's motion, we need to determine the position of the car at different points in time.
Given:
- Initial velocity (u) = 15 m/s
- Time taken to stop (t) = 6 s
To find the position of the car at different points in time, we can use the equation of motion:
s = ut + 0.5at^2
Since the car comes to a stop, the final velocity (v) would be 0 m/s. Thus, we can solve for acceleration (a):
v = u + at
0 = 15 + a(6)
-15 = 6a
a = -15/6
a = -2.5 m/s^2
Now that we have the acceleration, we can find the position of the car at different points in time.
Let's assume the car starts from position 0 (s = 0) at t = 0.
For t = 0 s:
s = 0
For t = 1 s:
s = ut + 0.5at^2
s = 15(1) + 0.5(-2.5)(1)^2
s = 15 - 1.25
s = 13.75 m
For t = 2 s:
s = ut + 0.5at^2
s = 15(2) + 0.5(-2.5)(2)^2
s = 30 - 5
s = 25 m
For t = 3 s:
s = ut + 0.5at^2
s = 15(3) + 0.5(-2.5)(3)^2
s = 45 - 11.25
s = 33.75 m
For t = 4 s:
s = ut + 0.5at^2
s = 15(4) + 0.5(-2.5)(4)^2
s = 60 - 20
s = 40 m
For t = 5 s:
s = ut + 0.5at^2
s = 15(5) + 0.5(-2.5)(5)^2
s = 75 - 31.25
s = 43.75 m
For t = 6 s:
s = ut + 0.5at^2
s = 15(6) + 0.5(-2.5)(6)^2
s = 90 - 45
s = 45 m
Now, we can plot these points on a graph with time (t) on the x-axis and position (s) on the y-axis:
(time t) | (position s)
----------------------------
0 | 0
1 | 13.75
2 | 25
3 | 33.75
4 | 40
5 | 43.75
6 | 45
Using these points, you can connect them with a line to represent the motion of the car until it comes to a stop at 6 seconds.
To draw a graph showing the results of a car's deceleration, we need to determine the car's position as a function of time.
1. First, we need to calculate the car's acceleration using Newton's second law of motion: F = ma, where F is the net force acting on the car, m is the mass of the car, and a is the acceleration.
The net force on the car is given by the braking force, which can be calculated using the equation: F = m * a, where m is the mass of the car and a is the acceleration.
Rearranging the formula, we get: a = F / m.
2. The deceleration is the negative value of acceleration, as the car is slowing down. So, set a as a negative value (-a).
3. Calculate the deceleration using the equation:
a = ((0 m/s) - (15 m/s)) / 6 s
Plugging in the values, we get:
a = (-15 m/s) / 6 s
a ≈ -2.5 m/s²
4. Now, we can use the formula for position as a function of time when acceleration is constant:
x = x₀ + v₀t + (1/2)at²
In this equation, x is the position at time t, x₀ is the initial position (0 m in this case as the car started to brake from rest), v₀ is the initial velocity (15 m/s), t is the time, and a is the acceleration.
5. Plug in the values and calculate the position at different time intervals. Here is a table of positions at different times:
| Time (s) | Position (m) |
|----------|--------------|
| 0 | 0 |
| 1 | 6.25 |
| 2 | 20 |
| 3 | 37.5 |
| 4 | 60 |
| 5 | 87.5 |
| 6 | 120 |
6. Plot these points on a graph with time (t) on the x-axis and position (x) on the y-axis. Connect the points with a smooth curve.
The resulting graph will show the position of the car as it decelerates over time. The initial position is at (0,0) and it gradually increases until it reaches a final position of (6, 120) as the car comes to a complete stop.