The acceleration of a bus is given by ax (t) = at, where a = 1.2 m/s3.If the bus's velocity at time t = 1.0s is 5.0 m/s, what is its velocity at time t = 2.0 s?
sorry. acceleration a is in m/s^2
and what the heck is ax(t) = at?
In any case, if a is constant, v(2) = v(1) + a
That's what acceleration is -- how much the velocity changes in a second.
a = ∆v/∆t where ∆t=1 in your question.
V = Vo + a*t.
5 = Vo + 1.2*1.0,
Vo = 3.8 m/s.
V = Vo + a*t. = 3.8 + 1.2*2 = __m/s.
To find the velocity of the bus at time t = 2.0 s, we need to integrate the acceleration function with respect to time from t = 1.0 s to t = 2.0 s.
Given that the acceleration function is ax(t) = at, with a = 1.2 m/s^3, we can integrate it to find the velocity function.
∫ ax(t) dt = ∫ at dt
Integrating both sides of the equation gives:
v(t) = ∫ at dt
Integrating with respect to t, we get:
v(t) = (1/2)at^2 + C
Where C is the constant of integration.
Now, let's find the value of C using the given information that the bus's velocity at t = 1.0 s is 5.0 m/s:
v(1) = (1/2)a(1)^2 + C = 5.0 m/s
Simplifying this equation yields:
(1/2)a + C = 5.0
Substituting the value of a = 1.2 m/s^3, we can solve for C:
(1/2)(1.2) + C = 5.0
0.6 + C = 5.0
C = 5.0 - 0.6
C = 4.4 m/s
Therefore, the constant of integration C is 4.4 m/s.
Now that we have the value of C, we can find the velocity at t = 2.0 s by substituting t = 2.0 s into the velocity function:
v(2) = (1/2)a(2)^2 + C
Substituting the values of a = 1.2 m/s^3 and C = 4.4 m/s, we get:
v(2) = (1/2)(1.2)(2)^2 + 4.4
v(2) = (1/2)(1.2)(4) + 4.4
v(2) = 2.4 + 4.4
v(2) = 6.8 m/s
Therefore, the velocity of the bus at time t = 2.0 s is 6.8 m/s.