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, we can use the equations of motion.
Let's denote the initial position of the particle as s₀ (which is unknown), the initial velocity as u (what we are trying to find), the acceleration as a (what we need to find), the time as t, and the distance traveled as s.
From the given information, we know that the particle covered 6m in the 4th second, and 7m in the 5th second. This means that the distances traveled between these time intervals are:
s₁ = 6m
s₂ = 7m
We also know that the time intervals between these distances are:
t₁ = 4s
t₂ = 5s
Let's start by finding the acceleration (a).
We can use the equation of motion: s = ut + 0.5at²
For the first interval (4th second):
s₁ = u(t₁) + 0.5a(t₁)²
6m = u(4s) + 0.5a(4s)²
6m = 4u + 8a
For the second interval (5th second):
s₂ = u(t₂) + 0.5a(t₂)²
7m = u(5s) + 0.5a(5s)²
7m = 5u + 12.5a
Now, we have two equations with two unknowns (u and a). We can solve these equations simultaneously to find the values of u and a.
Equation 1: 6m = 4u + 8a
Equation 2: 7m = 5u + 12.5a
We can solve these equations using any method of solving simultaneous equations, such as substitution or elimination.
Let's use the elimination method to solve for u:
Multiply Equation 1 by 5 and Equation 2 by 4 to eliminate the 'u' term:
(5 * Equation 1): 30m = 20u + 40a
(4 * Equation 2): 28m = 20u + 50a
Now, subtract Equation 2 from Equation 1:
30m - 28m = (20u + 40a) - (20u + 50a)
2m = -10a
a = -0.2 m/s² (acceleration)
To find the initial velocity (u), substitute the value of 'a' back into one of the equations (for example, Equation 1):
6m = 4u + 8a
6m = 4u + 8(-0.2m/s²)
6m = 4u - 1.6m/s²
Rearrange the equation to solve for 'u':
4u = 6m + 1.6m/s²
4u = 7.6m/s²
u = 1.9 m/s (initial velocity)
Therefore, the acceleration of the particle is -0.2 m/s², and its initial speed is 1.9 m/s.