The position of a particle moving on X-axis is given by
`x=At^3+Bt^2+Ct+D`.
The numerical values of A,B,C,D are 1,4,-2 and 5 respectively and I units are used. Find a. the dimensions of A,B, C and D b. the velocity of the particle at t=4s, the acceleration of he particle at t=4s, d. The average velocity during the interval t=0 to t=4s, the average accelerationduring the interval t=0 to t=4s.

Explanation

a. Dimensons of x, `At^3, Bt^2, Ct and D` must be indentica and in this case each is length. Thus,
`[At^2]=L, or [A] =LT^-3`
`[Bt^&2]=L, or [B] = LT^-2`
`[Ct]=L, or [C]=LT^-1`
and `[D]=L`
b. `x=At^3+Bt^2+Ct+D`
or, `v=(dx)/(dt)=3At^2+2Bt+C`
Thus at t=4s, the velocity `
=3(1/s^3) (16s^2)+(4m/s^2)(4s)+(-2 m/s)`
`=(48+32-2) m/s=78 m/s`
c. v=3At^2+2Bt+C`
or, `a=(dv)/(dt)=6At+2B`.
At=2,a=6(1 m/s^3)(4s)+2(4m/s^2)=32m/s^2`
d. `x=At^3+Bt^2+Ct+D`
Position at t=0 is x=D =m`
Position at t=4 s is
`(m/s^3)(64s^3)+(4m/s^2)(16s^2)-(2m/s)(4s)+5m`
=()64+64-8+5)m=125m`
Thus, the displacement during 0 to 4 s is
`125m-5m=120m`
Average velocity `=(120m)/(4s)=30 m/s`
e. `v=3At^2+2Bt+C`
Velocity at t=0 is C=-2 m/s,
Velocity at t=4s is =78 m/s
Average acceleration `=(v_2-v_1)/(t_2-t_1)=20 m/s^2`.