Question

The position of a body is given by x = At + 6Bt3, where x is in meters and t is in seconds. (a) What are the units of A and B? (b) What is the acceleration as a function of time? (c) What is the velocity and acceleration at t = 5.0 s? (d) What is the velocity as a function of time if x = At + Bt-3?

Answer 1

Answer:

(a)

Therefore the unit of A is m/s and the unit of B is m/s³.

(b)

Therefore the acceleration of the body is 36Bt.

(c)

Therefore the velocity = A+450B m/s

and acceleration =180B m/s²

(d)

Therefore the velocity of the function is ( $A-3Bt^{-4}$) m/s.

Step-by-step explanation:

Given that the position of a body is given by

$x= At+6B t^3$

(a)

x is in meters , so the unit of At and Bt³ are also m. t is in second.

At= meters

$\Rightarrow A=\frac{meters}{second}$

Bt³ = meters

$\Rightarrow B=\frac{meters}{second^3}$

Therefore the unit of A is m/s and the unit of B is m/s³.

(b)

The first order derivative of the position of a particle give the velocity of the particle.

The second order derivative of the position of a particle give the acceleration of the particle.

$x= At+6B t^3$

Differentiating with respect to t

$v=\frac{dx}{dt}=A+6.B.3t^2$

$\Rightarrow \frac{dx}{dt}=A+18Bt^2$

Again differentiating with respect to t

$a=\frac{d^2x}{dt^2}= 18.B.2t$

$\Rightarrow \frac{d^2x}{dt^2}= 36Bt$

Therefore the acceleration of the body is 36Bt.

(c)

$\frac{dx}{dt}|_{t=5.0}=A+18B.5^2$

            =A+450B

$\frac{d^2x}{dt^2}|_{t=5}= 36B .5$

           =180B

Therefore the velocity = A+450B m/s

and acceleration =180B m/s²

(d)

Given function is

$x=At+Bt^{-3}$

Differentiating with respect to t

$v=\frac{dx}{dt}=A+(-3)Bt^{-3-1}$

           $=A-3Bt^{-4}$

Therefore the velocity of the function is ( $A-3Bt^{-4}$) m/s.