Question

The acceleration of a bus is given by ax(t) = αt, where α = 1.2 m/s3. (a) If the bus’s velocity at time t = 1.0 s is 5.0 m/s, what is its velocity at time t = 2.0 s? (b) If the bus’s position at time t = 1.0 s is 6.0 m, what is its position at time t = 2.0 s? (c) Sketch ay-t, vy-t , and x-t graphs for the motion.

Answer 1

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).  $a_{y}-t$, $v_{y}-t$ and x-t graphs

Explanation:

Given that,

$\alha=1.2\ m/s^3$

Time t = 1.0 s

Velocity = 5.0

The Acceleration equation is

$a_{x(t)}=\alpha t$

We need to calculate the velocity

Using formula of acceleration

$a=\dfrac{dv}{dt}$

On integrating

$\int_{v_{0}}^{v}{dv}=\int_{0}^{t}{a dt}$

Put the value into the formula

$v-v_{0}=1.2\int_{0}^{t}{t dt}$

$v-v_{0}=0.6t^2$

$v=v_{0}+0.6t^2$

Put the value into the formula

$v_{0}=5.0-0.6\times(1.0)^2$

$v_{0}=4.4\ m/s$

We need to calculate the velocity at 2.0 sec

Put the value of initial velocity in the equation

$v=4.4+0.6\times(2.0)^2$

$v=6.8\ m/s$

(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

$v=\dfrac{dx}{dt}$

On integrating

$\int_{x_{0}}^{x}{dx}=\int_{0}^{t}{v dt}$

$x_{0}-x=\int_{0}^{t}{v_{0}dt}+\int_{0}^{t}{0.6 t^2}$

$x_{0}-x=v_{0}t+\dfrac{0.6}{3}t^3$

$x=x_{0}+v_{0}t+\dfrac{0.6}{3}t^3$

$x_{0}=6-4.4\times1-\dfrac{0.6}{3}\times1^3$

$x=1.4\ m$

The position at t = 2.0 s

$x=1.4+4.4\times2.0+\dfrac{0.6}{3}\times2^3$

$x=11.8\ m$

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).  $a_{y}-t$, $v_{y}-t$ and x-t graphs