A particle is travelling on a straight line with constant acceleration. It covers 6m in the 4th second and 7m in the 5th second. Fine its acceleration and initial speed.
To find the acceleration and initial speed of the particle, you can use the equations of motion for uniformly accelerated linear motion.
The equation we'll use is:
s = ut + (1/2)at^2
where:
- s is the displacement of the particle
- u is the initial velocity
- t is the time
- a is the acceleration
We have two pieces of information:
1) The particle covers 6m in the 4th second:
s = 6m
t = 4s
Plugging these values into the equation, we get:
6 = u(4) + (1/2)a(4)^2
6 = 4u + 8a --(1)
2) The particle covers 7m in the 5th second:
s = 7m
t = 5s
Plugging these values into the equation, we get:
7 = u(5) + (1/2)a(5)^2
7 = 5u + 12.5a --(2)
Now, we have a system of linear equations (equations 1 and 2) that we can solve to find the values of u (initial velocity) and a (acceleration).
To solve this system of equations, you can use several methods such as substitution or elimination. I'll use the substitution method here.
First, solve equation (1) for u in terms of a:
u = (6 - 8a) / 4 --(3)
Now substitute equation (3) into equation (2):
7 = 5[(6 - 8a) / 4] + 12.5a
Multiplying both sides by 4 to eliminate the fractions, we get:
28 = 5(6 - 8a) + 50a
Distribute and simplify:
28 = 30 - 40a + 50a
28 = 30 + 10a
10a = -2
a = -0.2 m/s^2
Now substitute this value of a back into equation (3) to find u:
u = (6 - 8(-0.2)) / 4
u = 6.8 / 4
u = 1.7 m/s
Therefore, the acceleration of the particle is -0.2 m/s^2 (negative sign indicating that it's decelerating), and the initial speed is 1.7 m/s.