Question

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

Answer 1

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"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:

$v(t) = \int a(t) dt = \int (\alpha t)dt=\frac{1}{2}\alpha t^2+C$ (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:

$C=v(t) - \frac{1}{2}\alpha t^2 = 5.09 - \frac{1}{2}(1.28)(1.13)^2=4.27 m/s$

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

$v(2.02)=\frac{1}{2}(1.28)(2.02)^2+4.27=6.88 m/s$

B) 11.2 m

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

$x(t) = \int v(t) dt = \int (\frac{1}{2}\alpha t^2 + C) dt = \frac{1}{3}\alpha t^3 + Ct +D$ (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:

$D = x(t) - \frac{1}{6}\alpha t^3 -Ct = 5.92-\frac{1}{6}(1.28)(1.13)^3 - (4.27)(1.13)=0.79 m$

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

$x(2.02)=\frac{1}{6}(1.28)(2.02)^3 + (4.27)(2.02) +0.79=11.2 m$