a)The figure below shows a velocity-time graph of a particle starts from rest and accelerates at 11/4metres per second ^2 for t1,seconds until it attains a velocity of 22m/sIt then maintains this velocity upto the 40th second.Breaks are applied bringing it uniformly to rest after t2.The particle covers 847 metres before it comes to rest.
Find;
i)The value of t1
ii)The deceleration time
b)The distance between two trains T1 and T2 travelling towards the same direction is 100metres.T1 is 60 metres long and travels at 80k/h while T2 is 140 metres long and travels at 60km/h.Find the time in seconds it takes T1 to completely overtake T2.
a) i) To find the value of t1, we need to find the time it takes for the particle to reach a velocity of 22 m/s. From the graph, we can see that the particle starts from rest and accelerates at 11/4 m/s^2. We can use the formula:
v = u + at
where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration, and t is the time. Rearranging the formula, we have:
t = (v - u) / a
Substituting the given values, we have:
t1 = (22 - 0) / (11/4) = 8 seconds
So, the value of t1 is 8 seconds.
ii) To find the deceleration time (t2), we need to find the time it takes for the particle to come to rest after the velocity of 22 m/s. From the graph, we can see that the particle maintains this velocity until the 40th second, and then breaks are applied to bring it uniformly to rest.
The distance covered by the particle before it comes to rest (s) is given as 847 meters. We can use the equation:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. In this case, we need to find the time when the velocity is 0 m/s, so u is 22 m/s and a is the deceleration.
Substituting the given values, we have:
847 = 22 * t2 - (1/2) * a * t2^2
Since the particle decelerates uniformly, the final velocity is 0 m/s. Rearranging the equation, we have:
(1/2) * a * t2^2 - 22 * t2 + 847 = 0
Solving this quadratic equation for t2, we find two possible values:
t2 ≈ 3.061 seconds or t2 ≈ 14.36 seconds
However, since the particle breaks after the 40th second, the deceleration time must be less than 40 seconds. Therefore, the value of t2 is 3.061 seconds.
b) To find the time it takes for train T1 to completely overtake train T2, we need to find the time it takes for T1 to cover a distance equal to the sum of the lengths of T2 and the distance between the two trains.
The distance between the two trains is 100 meters. Train T1 is 60 meters long and travels at 80 km/h, which is 80 * (1000/3600) = 22.22 m/s. Train T2 is 140 meters long and travels at 60 km/h, which is 60 * (1000/3600) = 16.67 m/s.
Let's denote the time it takes for T1 to completely overtake T2 as t. In this case, T1 covers a distance of 60 + 100 = 160 meters in time t, at a velocity of 22.22 m/s. T2 covers a distance of 140 meters in time t, at a velocity of 16.67 m/s.
Setting up the equation:
160 = 22.22 * t
140 = 16.67 * t
We can solve these equations simultaneously to find the value of t. Dividing the second equation by the first equation, we have:
140/160 = (16.67 * t) / (22.22 * t)
0.875 = 0.75
t = 0.75
Therefore, it takes T1 0.75 seconds to completely overtake T2.