the motion of a particle along a straight line is described by the function x=(2t-3)^2 where x is in metres and t is in seconds.
A)find the position ,veocity and acceleration at t=2 sec.
B) find the velocity of the particle at origin.
A) To find the position, velocity, and acceleration at t=2 seconds, we need to differentiate the given function with respect to time.
Given: x = (2t-3)^2
1. Position (x):
To find the position at t=2 sec, substitute t=2 into the function:
x = (2(2)-3)^2
x = (4-3)^2
x = 1^2
x = 1 meter
2. Velocity (v):
The velocity is the first derivative of the position function. Differentiating x with respect to t:
v = d/dt (x)
v = d/dt [(2t-3)^2]
Using the chain rule of differentiation, we get:
v = 2(2t-3) * d/dt (2t-3)
v = 2(2t-3) * 2
v = 4(2t-3)
Substituting t=2:
v = 4(2(2)-3)
v = 4(4-3)
v = 4 m/s
So, at t=2 sec, the position is 1 meter and the velocity is 4 m/s.
3. Acceleration (a):
The acceleration is the derivative of velocity. Differentiating v with respect to t:
a = d/dt (v)
a = d/dt [4(2t-3)]
Using the chain rule again:
a = 4 * d/dt (2t-3)
a = 4 * 2
a = 8 m/s^2
So, at t=2 sec, the position is 1 meter, velocity is 4 m/s, and acceleration is 8 m/s^2.
B) To find the velocity of the particle at the origin (t=0 sec), we can substitute t=0 into the velocity equation:
v = 4(2t-3)
v = 4(2(0)-3)
v = 4(-3)
v = -12 m/s
Therefore, the velocity of the particle at the origin (t=0 sec) is -12 m/s.
A) To find the position, velocity, and acceleration at t = 2 sec, we need to differentiate the given function with respect to time (t).
1) Position:
Given function: x = (2t - 3)^2
To find the position at t = 2 sec, substitute t = 2 into the function:
x = (2(2) - 3)^2
x = (4 - 3)^2
x = 1^2
x = 1 meter
So, the position of the particle at t = 2 sec is 1 meter.
2) Velocity:
To find the velocity, we need to differentiate the function x = (2t - 3)^2 with respect to t.
Differentiating the function:
dx/dt = 2(2t - 3)(2) [Applying the power rule of differentiation]
dx/dt = 4(2t - 3)
dx/dt = 8t - 12
To find the velocity at t = 2 sec, substitute t = 2 into the derivative:
v = 8(2) - 12
v = 16 - 12
v = 4 m/s
So, the velocity of the particle at t = 2 sec is 4 m/s.
3) Acceleration:
To find the acceleration, we need to differentiate the velocity function dx/dt = 8t - 12 with respect to t.
Differentiating the velocity function:
d^2x/dt^2 = 8
The constant coefficient 8 indicates that the acceleration is constant and does not depend on time.
So, the acceleration of the particle at t = 2 sec is 8 m/s^2.
B) To find the velocity of the particle at the origin, we need to substitute x = 0 into the function x = (2t - 3)^2 and solve for t.
Given function: x = (2t - 3)^2
Substituting x = 0:
0 = (2t - 3)^2
Taking the square root of both sides:
√0 = √(2t - 3)^2
0 = ±(2t - 3)
Solving for t, we get two cases:
1) 2t - 3 = 0
2t = 3
t = 3/2 = 1.5 seconds
2) -(2t - 3) = 0
-2t + 3 = 0
2t = 3
t = 3/2 = 1.5 seconds
Therefore, at the origin (x = 0), the velocity of the particle occurs at t = 1.5 seconds.
qwsfd
A) position: x(2)= (2*2-3)^2= 1m
v = dx/dt = 2(2t-3)*2
so v(2)= 4 m/s
a = dv/dt = 8 m/s^2
B) v at origin: x=0=(2t-3)^2
so t=3/2
at t=3/2, v = 4(2*3/2-3)= 0