Question

The acceleration of a bus is given by ax(t)=αt, where α = 1.28 m/s3 is a constant.

Answer 1

The question deals with calculating the time-dependent acceleration of a bus using the equation ax(t)=αt, applicable to Physics (specifically mechanics), and is suitable for a high school level.

Step-by-step explanation:

The question pertains to the acceleration of an object, specifically a bus, which makes this a Physics problem. Acceleration is defined as the change in velocity over time and is a vector quantity having both magnitude and direction. Given the equation ax(t)=αt, where α is a constant equal to 1.28 m/s3, we can recognize that the acceleration is linearly increasing over time because for each second, the acceleration increases by 1.28 m/s2. This kind of problem typically appears in high school Physics.

For instance, to calculate the acceleration of a bus at a specific time point, we simply multiply the constant α by the time t at which we want to know the acceleration. So if we wanted to know the acceleration at t=15 s, we would calculate it as ax(15s) = 1.28 m/s3 × 15 s = 19.2 m/s2. This demonstrates that as time progresses, the acceleration value increases.

Answer 2

The text looks incomplete. Here the complete question found on google:

"The acceleration of a bus is given by ax(t)=αt, where α = 1.28m/s3 is a constant. Part A If the bus's velocity at time t1 = 1.13s is 5.09m/s , what is its velocity at time t2 = 2.02s ? If the bus's position at time t1 = 1.13s is 5.92m , what is its position at time t2 = 2.02s ?"

A) 6.88 m/s

The velocity of the bus can be found by integrating the acceleration. Therefore:

`v(t) = \int a(t) dt = \int (\alpha t)dt=(1)/(2)\alpha t^2+C` (1)

where

`\alpha = 1.28 m/s^3`

C is a constant term

We know that at
`t_1 = 1.13 s`, the velocity is
`v=5.09 m/s`. Substituting these values into (1), we can find the exact value of C:

`C=v(t) - (1)/(2)\alpha t^2 = 5.09 - (1)/(2)(1.28)(1.13)^2=4.27 m/s`

So now we can find the velocity at time
`t_2 = 2.02 s`:

`v(2.02)=(1)/(2)(1.28)(2.02)^2+4.27=6.88 m/s`

B) 11.2 m

To find the position, we need to integrate the velocity:

`x(t) = \int v(t) dt = \int ((1)/(2)\alpha t^2 + C) dt = (1)/(3)\alpha t^3 + Ct +D` (2)

where D is another constant term.

We know that at
`t_1 = 1.13 s`, the position is
`v=5.92 m`. Substituting these values into (2), we can find the exact value of D:

`D = x(t) - (1)/(6)\alpha t^3 -Ct = 5.92-(1)/(6)(1.28)(1.13)^3 - (4.27)(1.13)=0.79 m`

And so now we can find the position at time
`t_2 = 2.02 s` using eq.(2):

`x(2.02)=(1)/(6)(1.28)(2.02)^3 + (4.27)(2.02) +0.79=11.2 m`