Question

Only for Physics specialists..

Instantaneous velocity is given as.. v = 5e^t m/s.

Find average acceleration and average Velocity of the particle between the instants t1 = 2 sec to t2 = 4 sec.

(Given at t=0 , x = 0)


Need solution​

Answer 1

Answer:

The quantity that tells us how fast an object is moving anywhere along its path is theinstantaneous velocity, usually called simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position x as a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation is –v=x(t2)−x(t1)t2−t1v–=x(t2)−x(t1)t2−t1. To find the instantaneous velocity at any position, we let t1=tt1=t and t2=t+Δtt2=t+Δt. After inserting these expressions into the equation for the average velocity and taking the limit as Δt→0Δt→0, we find the expression for the instantaneous velocity:

v(t)=limΔt→0x(t+Δt)−x(t)Δt=dx(t)dt.v(t)=limΔt→0x(t+Δt)−x(t)Δt=dx(t)dt.

Instantaneous Velocity

The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of xwith respect to t:

v(t)=ddtx(t).v(t)=ddtx(t).

Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point t0t0 is the rate of change of the position function, which is the slope of the position function x(t)x(t) at t0t0. (Figure) shows how the average velocity 

Answer 2

x = 5∫e^t dt

x = { 5 e^t } t1 = 2 , t2 = 4

x = 5 e^4 - 5 e^2

x = 5e^2 { e^2 - 1 }

V avg. = 5e^2 { e^2 - 1 } / 6

V avg. = 5/6 { e^2 ( e^2 - 1 ) }

v = 5 e^t m / s

Acc. avg. = 5 e^4 - 5e^2 / 2

Acc. avg. = 5/2 { e^2 ( e^2 - 1 ) }