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

(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

Step-by-step 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=(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*(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*(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=(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+(0.6)/(3)t^3`

`x=x_(0)+v_(0)t+(0.6)/(3)t^3`

`x_(0)=6-4.4*1-(0.6)/(3)*1^3`

`x=1.4\ m`

The position at t = 2.0 s

`x=1.4+4.4*2.0+(0.6)/(3)*2^3`

`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