A particle travelling along a straight line starts with a
velociry ofu metres per second and its speed increases
per second by rl metres. The particle travels 10 metres
in 2 seconds and 28 metres in 4 seconds. Find u and a.
(lf the distance travelled in I seconds is s metres. then
s=ut+1/2at^2
v = u + a t
s = u t + (1/2) a t^2
10 = u(2)+ + (1/2) a (2)^2
or 10 = 2 u + 2 a
28 = u(4) + (1/2) a (16)
or
28 = 4 u + 8 a
20 = 4 u + 4 a
28 = 4 u + 8 a
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8 = 4 a etc
To find the initial velocity (u) and the acceleration (a) of the particle, we can use the equation of motion:
s = ut + (1/2)at^2
Given that the particle travels 10 meters in 2 seconds, we can substitute these values into the equation to get:
10 = 2u + (1/2)a(2^2)
10 = 2u + 2a
Equation 1: 2u + 2a = 10
Similarly, for the particle traveling 28 meters in 4 seconds, we can substitute those values into the equation to get:
28 = 4u + (1/2)a(4^2)
28 = 4u + 8a
Equation 2: 4u + 8a = 28
Now we have a system of linear equations that we can solve to find the values of u and a.
To do this, we can multiply Equation 1 by 2 and subtract it from Equation 2:
(4u + 8a) - 2(2u + 2a) = 28 - 2(10)
4u + 8a - 4u - 4a = 28 - 20
4a = 8
a = 2
Substituting this value of a back into Equation 1, we get:
2u + 2(2) = 10
2u + 4 = 10
2u = 6
u = 3
Therefore, the initial velocity (u) of the particle is 3 m/s, and the acceleration (a) is 2 m/s^2.