Q5 -f-Q5- The position of an object moving along a straight line is given by x - 3 ndash- 2t2 - 3t3 where x is in meters and t in seconds- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-a- Derive the expressions for the velocity and acceleration of the object as a function of time- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-b- Find the average velocity between t - 2 s and t - 4 s - between t - 0s and t - 45 - between t - 1 s and t - 3 s- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-c- What is th instantaneous velocity at t - 2 s- At t - 5 s - nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-d- At what time -s- is - are the instantaneous velocities zero- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-e- When does the instantaneous velocity have a maximum or a minimum value- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-f- Find the average acceleration between t - 2 s and t - 5 s- between t - 1 s and t - 3 s nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-g- When is the instantaneous acceleration of the object zero- nbsp-

Dear Student f. x= #160;3 2t2+3t3v=dxdt= 4t+9t2aavg= #8710;v #8710;tv2= 8+36=28v5= 20+225=205aavg= #8710;v #8710;t=v5 v25 2=205 283=59 #160;m/s2v1= ...

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Byju's Answer

Standard XII

Physics

Relating Circular Motion to SHM

Q5 (f) Q5...

Question

Q5 (f)

Q5. The position of an object moving along a straight line is given by x = 3 – 2t² + 3t³ where x is in meters and t in seconds.

(a) Derive the expressions for the velocity and acceleration of the object as a function of time.

(b) Find the average velocity between t = 2 s and t = 4 s ; between t = 0s and t = 45 ; between t = 1 s and t = 3 s.

(c) What is th instantaneous velocity at t = 2 s? At t = 5 s ?

(d) At what time (s) is / are the instantaneous velocities zero?

(e) When does the instantaneous velocity have a maximum or a minimum value?

(f) Find the average acceleration between t = 2 s and t = 5 s; between t = 1 s and t = 3 s

(g) When is the instantaneous acceleration of the object zero?

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Solution

Dear Student

f.
$x = 3 - 2 t^{2} + 3 t^{3} v = \frac{d x}{d t} = - 4 t + 9 t^{2} a_{a v g} = \frac{∆ v}{∆ t} v_{2} = - 8 + 36 = 28 v_{5} = - 20 + 225 = 205 a_{a v g} = \frac{∆ v}{∆ t} = \frac{v_{5} - v_{2}}{5 - 2} = \frac{205 - 28}{3} = 59 m / s^{2} v_{1} = - 4 + 9 = 5 v_{3} = - 12 + 81 = 69 a_{a v g} = \frac{∆ v}{∆ t} = \frac{v_{3} - v_{1}}{3 - 1} = \frac{69 - 5}{2} = 32 m / s^{2}$
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Dear Student f. x= 3-2t2+3t3v=dxdt=-4t+9t2aavg=∆v∆tv2=-8+36=28v5=-20+225=205aavg=∆v∆t=v5-v25-2=205-283=59 m/s2v1=-4+9=5v3=-12+81=69aavg=∆v∆t=v3-v13-1=69-52=32 m/s2 Regards