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). [tex]a_{y}-t[/tex], [tex]v_{y}-t[/tex] and x-t graphs
Explanation:
Given that,
[tex]\alha=1.2\ m/s^3[/tex]
Time t = 1.0 s
Velocity = 5.0
The Acceleration equation is
[tex]a_{x(t)}=\alpha t[/tex]
We need to calculate the velocity
Using formula of acceleration
[tex]a=\dfrac{dv}{dt}[/tex]
On integrating
[tex]\int_{v_{0}}^{v}{dv}=\int_{0}^{t}{a dt}[/tex]
Put the value into the formula
[tex]v-v_{0}=1.2\int_{0}^{t}{t dt}[/tex]
[tex]v-v_{0}=0.6t^2[/tex]
[tex]v=v_{0}+0.6t^2[/tex]
Put the value into the formula
[tex]v_{0}=5.0-0.6\times(1.0)^2[/tex]
[tex]v_{0}=4.4\ m/s[/tex]
We need to calculate the velocity at 2.0 sec
Put the value of initial velocity in the equation
[tex]v=4.4+0.6\times(2.0)^2[/tex]
[tex]v=6.8\ m/s[/tex]
(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
[tex]v=\dfrac{dx}{dt}[/tex]
On integrating
[tex]\int_{x_{0}}^{x}{dx}=\int_{0}^{t}{v dt}[/tex]
[tex]x_{0}-x=\int_{0}^{t}{v_{0}dt}+\int_{0}^{t}{0.6 t^2}[/tex]
[tex]x_{0}-x=v_{0}t+\dfrac{0.6}{3}t^3[/tex]
[tex]x=x_{0}+v_{0}t+\dfrac{0.6}{3}t^3[/tex]
[tex]x_{0}=6-4.4\times1-\dfrac{0.6}{3}\times1^3[/tex]
[tex]x=1.4\ m[/tex]
The position at t = 2.0 s
[tex]x=1.4+4.4\times2.0+\dfrac{0.6}{3}\times2^3[/tex]
[tex]x=11.8\ m[/tex]
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). [tex]a_{y}-t[/tex], [tex]v_{y}-t[/tex] and x-t graphs
