A particle is travelling in a straight line with constant acceleration. It covers 6m in the 4th second and 7m in the 5th second,find its acceleration and the initial speed.
If its speed at the start of the 4th second is v, then t seconds later, the distance is s = vt + 1/2 at^2
So now you know that
s(1)-s(0) = 6
s(2)-s(1) = 7
That is,
v + 1/2 a = 6
(2v + 2a) - (v + 1/2 a) = 7
--------------------------------
2v+a = 12
2v+3a = 14
----------------
a = 1
v = 11/2
To find the acceleration and initial speed of a particle traveling in a straight line with constant acceleration, we can use the equations of motion.
Let's break down the given information:
1. The particle covers 6 meters in the 4th second.
2. The particle covers 7 meters in the 5th second.
To solve for the acceleration and initial speed, we can follow these steps:
Step 1: Find the acceleration (a):
We can use the equation of motion relating displacement (s), initial velocity (u), time (t), and acceleration (a):
s = ut + (1/2)at^2
For the 4th second:
s1 = 6m
t1 = 4s
6 = u(4) + (1/2)a(4)^2
6 = 4u + 8a [Equation 1]
For the 5th second:
s2 = 7m
t2 = 5s
7 = u(5) + (1/2)a(5)^2
7 = 5u + 12.5a [Equation 2]
Step 2: Solve the system of equations:
To solve the system of equations (Equation 1 and Equation 2), we'll need to eliminate u. Subtracting Equation 1 from Equation 2 will allow us to do that:
(7 = 5u + 12.5a) - (6 = 4u + 8a)
1 = 5u - 4u + 12.5a - 8a
1 = u + 4.5a
u = 1 - 4.5a [Equation 3]
Step 3: Substitute the value of u in Equation 3 back into Equation 1:
Using Equation 3, we can substitute the value of u into Equation 1:
6 = 4u + 8a
6 = 4(1 - 4.5a) + 8a
6 = 4 - 18a + 8a
6 = 4 - 10a
10a = 4 - 6
10a = -2
a = -0.2 m/s^2
Step 4: Substitute the value of a into Equation 3 to find the initial velocity (u):
Using the value of a, we can substitute it into Equation 3:
u = 1 - 4.5a
u = 1 - 4.5(-0.2)
u = 1 + 0.9
u = 1.9 m/s
Therefore, the acceleration of the particle is -0.2 m/s^2, and the initial speed is 1.9 m/s.