The velocity time graph shows the movement of a particle alobg a straight line. The particle passes a point, A, when t=0. It maontains a constant spee, u ms -1, for T seconds and then decelerates uniformly before coming to rest at B. If the average speed for the qhole movement is 40 ms, calculate the valie of u. Given that the deceleration betweeen T second and 4T second is 5 ms calculate the value of T.
To find the value of "u," we can use the equation of motion for uniform acceleration, which is:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the particle maintains a constant speed of "u" m/s for "T" seconds, so its final velocity at time "T" is also "u" m/s. This means that the acceleration during this time period is 0.
At time "4T" seconds, the particle comes to rest, so its final velocity is 0 m/s. We are given that the deceleration between "T" and "4T" seconds is 5 m/s.
Applying the equation of motion to these intervals:
At time "T" seconds:
0 = u + 0 * T
0 = u
Therefore, the value of "u" is 0 m/s.
At time "4T" seconds:
0 = u + (-5) * 3T
0 = u - 15T
Since we found that "u" is 0 m/s, this equation simplifies to:
0 = -15T
Since the time "T" cannot be zero (otherwise, there would be no movement), we can solve for "T" as follows:
0 = -15T
T = 0 / -15
T = 0
Therefore, the value of "T" is 0 seconds.
To find the value of u, we need to use the concept of average speed.
Average Speed = Total Distance / Total Time
First, let's determine the distance traveled during the initial constant speed phase, from t = 0 to t = T. Since the speed is constant, the distance is given by:
Distance = Speed × Time
Distance = u × T
Next, let's determine the distance traveled during the deceleration phase, from t = T to t = 4T. To find the distance, we need to calculate the area under the velocity-time graph during this phase. The graph is a triangle, with the base equal to T seconds and the height equal to the deceleration rate (5 m/s^2). The area of a triangle is given by:
Area = 0.5 × Base × Height
Distance = 0.5 × T × 5T
Distance = 2.5T^2
Finally, let's find the total distance traveled for the whole movement. The particle travels the distance from A to B during the constant speed phase (u × T), plus the distance traveled during the deceleration phase (2.5T^2). Therefore:
Total Distance = u × T + 2.5T^2
Given that the average speed for the whole movement is 40 m/s, we can set up the following equation:
Average Speed = Total Distance / Total Time
40 = (u × T + 2.5T^2) / (T + 3T)
40 = (u × T + 2.5T^2) / 4T
Simplifying further:
40 = (u + 2.5T) / 4
Now, let's solve for u:
u + 2.5T = 160
u = 160 - 2.5T
To solve for T, we can use the information about the deceleration phase. We know that the deceleration between T seconds and 4T seconds is 5 m/s. From the graph, we can see that the change in velocity during this period is u - 0 (initial velocity). Using the equation:
Acceleration = Change in Velocity / Time
-5 = (0 - u) / (4T - T)
-5 = -u / (3T)
Simplifying further:
u = 15T
We can substitute this expression for u into the previous equation:
15T = 160 - 2.5T
Now, solve for T:
15T + 2.5T = 160
17.5T = 160
T = 9.14 seconds (rounded to two decimal places)
Therefore, the value of u is:
u = 15T = 15 × 9.14 = 137.1 m/s (rounded to one decimal place)