The acceleration am/s^2 of a particle moving in a straight line is given by a = 18t – 4, where t is time in seconds. The initial velocity of the particle is 2 m/s
a) Find the expression for velocity in terms of t
b) Determine the time when the velocity is again 2m/s
I hope you know a little calculus.
dv/dt=a
v= INT a dt=9t^2-4t + vo
2= 9t^2-4t + 2
0=t(9t-4)
t= 4/9 sec
a) To find the expression for velocity in terms of time (t), we need to integrate the acceleration function with respect to time.
The given acceleration function is a = 18t - 4.
To integrate this, we use the power rule of integration, which states that the integral of t^n with respect to t is (1/(n+1)) * t^(n+1).
Integrating the given acceleration function, we get:
v = ∫(18t - 4) dt
v = (18/2) * t^2 - 4t + C
v = 9t^2 - 4t + C
Where C is the constant of integration.
Since the initial velocity of the particle is given as 2 m/s, we can substitute that into the expression:
2 = 9t^2 - 4t + C
Now we can solve for C by plugging in a known value of t, such as t = 0:
2 = 9(0)^2 - 4(0) + C
2 = C
So the expression for velocity in terms of t is:
v = 9t^2 - 4t + 2
b) To determine the time when the velocity is again 2 m/s, we can equate the expression for velocity (v) with 2 and solve for t:
2 = 9t^2 - 4t + 2
Subtracting 2 from both sides gives:
0 = 9t^2 - 4t
Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula. Factoring, we get:
0 = t(9t - 4)
Setting each factor equal to zero:
t = 0
9t - 4 = 0
Solving the second equation gives:
9t = 4
t = 4/9
So the time when the velocity is again 2 m/s is t = 4/9 seconds.