The acceleration of a bus is given by a_x(t)=At, where A= 1.21 is a constant.
A)if the bus's velocity at time t_1= 1.14s is 5.03m/s, what is its velocity at time t_2= 2.04 ?
B)if the bus's velocity at time t_1= 1.14s is 6.02m/s, what is its velocity at time t_2= 2.04 ?
1.73
To find the velocity of the bus at a given time, we can integrate the acceleration function with respect to time. In this case, the acceleration function is given by a_x(t) = At, where A = 1.21 is a constant.
A) To find the velocity at time t_2 = 2.04s, given the velocity at time t_1 = 1.14s is 5.03m/s:
1. Integrate the acceleration function with respect to time:
∫ a_x(t) dt = ∫ Atdt
2. Apply the power rule of integration:
∫ Atdt = (1/2)At^2 + C, where C is the constant of integration.
3. Evaluate the integral at t_2 and t_1:
v(t) = (1/2)At^2 + C
v(t_2) = (1/2)A(t_2)^2 + C
v(t_1) = (1/2)A(t_1)^2 + C
4. Substitute the known values:
5.03 = (1/2)A(1.14)^2 + C
Solve the equation for C.
5. Substitute the value of C into the equation for v(t):
v(t_2) = (1/2)A(t_2)^2 + C
Calculate the velocity at t_2.
B) To find the velocity at time t_2 = 2.04s, given the velocity at time t_1 = 1.14s is 6.02m/s:
1. Follow steps 1-4 from part A to find the value of C.
2. Substitute the value of C into the equation for v(t):
v(t_2) = (1/2)A(t_2)^2 + C
Calculate the velocity at t_2.
A) The velocity change is the integral of the acceleration vs time
change in V = INTEGRAL 1.21 t dt
from t = 1.14 to t = 2.04 s
= 1.21[(2.04)^2 - (1.14)^2]/2
B) Same approach; same velocity change; different starting velocity