A body moved on x-axis at time t secs, the displacement of the particle from the origin is x-metre and the velocity of the particle is vm/s when t=0, s=0 and v=0. 1) given that the particle moves at a constant acceleration and that v=1 when t=4, find a)the value of acceleration b) the value of x when t=4 2) given instead that the acceleration of the particle at time t secs is -3/8(t-2) squared m/s squared. A) show that v=1 when t=4 b) find x when t=4
To solve these problems, we will use the equations of motion under constant acceleration. The four equations of motion are:
1) v = u + at
2) s = ut + (1/2)at^2
3) v^2 = u^2 + 2as
4) s = ((u + v) / 2) x t
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = displacement
1) Given that v = 1 when t = 4, we need to find the acceleration (a) and the displacement (x) when t = 4.
a) We know that v = u + at. Since u = 0 (initial velocity) and v = 1, we can plug in these values to the equation to solve for a:
1 = 0 + a*4
1 = 4a
a = 1/4
Therefore, the acceleration is 1/4 m/s².
b) To find the displacement (x) when t = 4, we can use equation 2:
s = ut + (1/2)at²
s = 0 + (1/2)(1/4)(4²)
s = (1/2)(1/4)(16)
s = 2
Therefore, the displacement (x) when t = 4 is 2 meters.
2) Given that the acceleration at time t is -3/8(t-2)² m/s², we need to show that v = 1 when t = 4 and find the displacement (x) when t = 4.
a) To show that v = 1 when t = 4, we can integrate the acceleration function with respect to time to find the velocity function:
∫(-3/8(t-2)²) dt = -3/8 * ∫(t-2)² dt
Simplifying and integrating, we get:
= -3/8 * ∫(t² - 4t + 4) dt
= -3/8 * (t³/3 - 2t² + 4t) + C
Evaluating this equation at t = 4:
v = -3/8 * (4³/3 - 2(4)² + 4(4)) + C
v = -3/8 * (64/3 - 32 + 16) + C
v = -3/8 * (64/3 - 16) + C
v = -3/8 * (64/3 - 16/1) + C
v = -3/8 * (64/3 - 48/3) + C
v = -3/8 * 16/3 + C
v = -2 + C
Since it is given that v = 1 when t = 4, we can set v = 1 and solve for C:
1 = -2 + C
C = 1 + 2
C = 3
Therefore, the equation for velocity (v) is v = -2t + 3.
b) To find the displacement (x) when t = 4, we can integrate the velocity function with respect to time to find the displacement function:
∫(-2t + 3) dt = -2 * ∫t dt + 3 * ∫dt
Simplifying and integrating, we get:
= -2 * (t²/2) + 3t + C
= -t² + 3t + C
Evaluating this equation at t = 4:
x = -(4)² + 3(4) + C
x = -16 + 12 + C
x = -4 + C
Since no initial value is given for displacement, we cannot determine the value of C, and thus we cannot find x when t = 4.