Answer:
(a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c).
,
and x-t graphs
Explanation:
Given that,
![\alha=1.2\ m/s^3](https://tex.z-dn.net/?f=%5Calha%3D1.2%5C%20m%2Fs%5E3)
Time t = 1.0 s
Velocity = 5.0
The Acceleration equation is
![a_{x(t)}=\alpha t](https://tex.z-dn.net/?f=a_%7Bx%28t%29%7D%3D%5Calpha%20t)
We need to calculate the velocity
Using formula of acceleration
![a=\dfrac{dv}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bdv%7D%7Bdt%7D)
On integrating
![\int_{v_{0}}^{v}{dv}=\int_{0}^{t}{a dt}](https://tex.z-dn.net/?f=%5Cint_%7Bv_%7B0%7D%7D%5E%7Bv%7D%7Bdv%7D%3D%5Cint_%7B0%7D%5E%7Bt%7D%7Ba%20dt%7D)
Put the value into the formula
![v-v_{0}=1.2\int_{0}^{t}{t dt}](https://tex.z-dn.net/?f=v-v_%7B0%7D%3D1.2%5Cint_%7B0%7D%5E%7Bt%7D%7Bt%20dt%7D)
![v-v_{0}=0.6t^2](https://tex.z-dn.net/?f=v-v_%7B0%7D%3D0.6t%5E2)
![v=v_{0}+0.6t^2](https://tex.z-dn.net/?f=v%3Dv_%7B0%7D%2B0.6t%5E2)
Put the value into the formula
![v_{0}=5.0-0.6\times(1.0)^2](https://tex.z-dn.net/?f=v_%7B0%7D%3D5.0-0.6%5Ctimes%281.0%29%5E2)
![v_{0}=4.4\ m/s](https://tex.z-dn.net/?f=v_%7B0%7D%3D4.4%5C%20m%2Fs)
We need to calculate the velocity at 2.0 sec
Put the value of initial velocity in the equation
![v=4.4+0.6\times(2.0)^2](https://tex.z-dn.net/?f=v%3D4.4%2B0.6%5Ctimes%282.0%29%5E2)
![v=6.8\ m/s](https://tex.z-dn.net/?f=v%3D6.8%5C%20m%2Fs)
(b). If the bus’s position at time t = 1.0 s is 6.0 m,
We need to calculate the position
Using formula of velocity
![v=\dfrac{dx}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bdx%7D%7Bdt%7D)
On integrating
![\int_{x_{0}}^{x}{dx}=\int_{0}^{t}{v dt}](https://tex.z-dn.net/?f=%5Cint_%7Bx_%7B0%7D%7D%5E%7Bx%7D%7Bdx%7D%3D%5Cint_%7B0%7D%5E%7Bt%7D%7Bv%20dt%7D)
![x_{0}-x=\int_{0}^{t}{v_{0}dt}+\int_{0}^{t}{0.6 t^2}](https://tex.z-dn.net/?f=x_%7B0%7D-x%3D%5Cint_%7B0%7D%5E%7Bt%7D%7Bv_%7B0%7Ddt%7D%2B%5Cint_%7B0%7D%5E%7Bt%7D%7B0.6%20t%5E2%7D)
![x_{0}-x=v_{0}t+\dfrac{0.6}{3}t^3](https://tex.z-dn.net/?f=x_%7B0%7D-x%3Dv_%7B0%7Dt%2B%5Cdfrac%7B0.6%7D%7B3%7Dt%5E3)
![x=x_{0}+v_{0}t+\dfrac{0.6}{3}t^3](https://tex.z-dn.net/?f=x%3Dx_%7B0%7D%2Bv_%7B0%7Dt%2B%5Cdfrac%7B0.6%7D%7B3%7Dt%5E3)
![x_{0}=6-4.4\times1-\dfrac{0.6}{3}\times1^3](https://tex.z-dn.net/?f=x_%7B0%7D%3D6-4.4%5Ctimes1-%5Cdfrac%7B0.6%7D%7B3%7D%5Ctimes1%5E3)
![x=1.4\ m](https://tex.z-dn.net/?f=x%3D1.4%5C%20m)
The position at t = 2.0 s
![x=1.4+4.4\times2.0+\dfrac{0.6}{3}\times2^3](https://tex.z-dn.net/?f=x%3D1.4%2B4.4%5Ctimes2.0%2B%5Cdfrac%7B0.6%7D%7B3%7D%5Ctimes2%5E3)
![x=11.8\ m](https://tex.z-dn.net/?f=x%3D11.8%5C%20m)
Hence, (a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c).
,
and x-t graphs