Question

Q. A particle is moving with a uniform acceleration 4m/s2 for time 2 seconds and then 5m/s2 for 3
seconds. What is the average acceleration of the particle during motion
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Answer 1

Answer:

6 sec

Explanation:

There are 3 cases

Case 1 : ascending with uniform accrltn

initial velocity u=0m/s , a= 4m/s² , time =t₁sec , final velocity =vm/s

distance covered in t₁ = s₁ m

v₁=0+4t₁ , 2xs₁x4+0²=v²⇒s₁=v²/8

case 2 : moving with uniform velocity v

since uniform velocity therefore no acceleration

initial velocity=vm/s , a=0m/s² , time =tsec. distance int=s₂ m final velocity = v m/s

s₂=vt

case3 : ascending with retardation

initial velocity =vm/s , time taken =t₂sec , distance travelled =s₃m

accleration =−4m/s² final velocity =0m/s

2x−4xs₃+v²=0² ⇒s₃=v²/8 , 0=v−4t₂⇒t₂=v/4

now A/Q

total time =t₁+t+t₂=8⇒v/4+t+v/4 ⇒t=8−v/2..........(1)

total distance =s₁+s₂+s₃=28 ⇒v²/8+v²/8+vt=28⇒v²/4+4vt=112

⇒substituting value of t from (1) gives quadratic equation in terms of v

v²−32v+112=0

on solving we get v=4m/s and v=28m/s

t=8−4/2=6sec and t=8−28/2=−6secas time can't be negative

true ans is v=4m/s and t=6secs

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Answer 2

Answer:

4.6 ms⁻²

Explanation:

The particle is moving with initial acceleration(a₁) of 4 ms⁻² for time(t₁) = 2 s.

Final velocity(v₁) after 2 s is given by the following equation.

v₁ = u₁ + a₁t₁ = 0 + 4×2 = 8 ms⁻¹. (∵ Initial velocity u₁= 0 ms⁻¹.)

Now, the particle further moves with changed acceleration(a₂) of 5 ms⁻² for time(t₂) = 3 s.

v₂ = u₂ + a₂t₂ = 8 + 5×3 = 23 ms⁻¹. (∵ Initial velocity u₂ = v₁ = 8 ms⁻¹)

Average acceleration = (Total change in velocity)/(Total time taken)

                                    = (v₂ - u₁)/(t₁+t₂) = (23 - 0)/(2 + 3) = 23/5 = 4.6 ms⁻². (Ans)