erPSE10 25 OP010 An object moves along the x axis according to the equation x3.4

Question

erPSE10 25 OP010 An object moves along the x axis according to the equation x3.45t2 2.00t +3.00, where x is in meter s and t is in seconds S. (a) Determine the average speed between t 2.90 s and t- 4.90 s. m/s (b) Determine the instantaneous speed at t 2.90 s. m/s Determine the instantaneous speed at t- 4.90 s (c) Determine the average acceleration between t2.90 s and t-4.90 s (d) Determine the instantaneous acceleration at t-2.90 s m/s m/s2 m/s2 Determine the instantaneous acceleration at t 4.90 s. m/s2 (e) At what time is the object at rest?

Explanation / Answer

Part B.

Instantaneous speed is given by:

V = dx/dt

V = d(3.45t^2 - 2t + 3)/dt

V = 2*3.45*t - 2*1 + 0

V = 6.90*t - 2

At t = 2.90 sec

V1 = 6.90*2.90 - 2 = 18.01 m/sec

at t = 4.90 sec

V2 = 6.90*4.90 - 2 = 31.81 m/sec

Part C.

Average acceleration is given by:

a = (Vf - Vi)/(tf - ti)

Vf = Velocity at 4.90 sec = 31.81 m/sec

Vi = Velocity at 2.90 sec = 18.01 m/sec

So,

a = (31.81 - 18.01)/(4.90 - 2.90)

a_avg = 6.90 m/sec^2

Part D.

instantaneous acceleration is

a = dV/dt

a = d(6.90*t - 2)/dt

a = 6.90*1 - 0 = 6.90 m/sec^2

Since above expression does not depend on time, So acceleration will be constant at all time

at t = 2.90 sec, acceleration = 6.90 m/sec^2

at t = 4.90 sec, acceleration = 6.90 m/sec^2

Part E.

Object will be at rest when V = 0

V = 6.90*t - 2 = 0

6.90*t = 2

t = 2/6.90

t = 0.29 sec

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