Given:

The time is t=2s .

The distance equation is s=(2t3) m .

Write the distance equation.

s=(2t3) m

Here, average velocity is υavg , change in distance is Δs and change in time is Δt .

Write the expression velocity.

υ=dsdt (I).

Here, velocity is υ and rate of change of distance s with respect to time (t) is dsdt .

Conclusion:

Substitute (2t3) m for s in Equation (I).

υ=d dt(2t3)=6t2 m/s (II).

Substitute 2 s for t in Equation (II).

υ=6t2 m/s=6(2)2=24 m/s 

Thus, the velocity when time is t=2s is 24 m/s .

Answer:

The acceleration a when distance s=1 m 25 m/s2 .

Given:

The distance is s=1 m .

The velocity equation is υ=(5 s) m/s .

Write the velocity equation.

υ=(5 s) m/s

Write the expression acceleration.

a=υdυds (I).

Here, velocity is υ , acceleration is a and rate of change of velocity υ with respect to distance (s) is dυds .

Conclusion:

Substitute (5 s) m/s for υ in Equation (I).

a=υdυds=(5 s) dds(5 s)=5s(5)=25 s (II).

Substitute 1 m for s in Equation (II).

a=25 s=25(1)=25 m/s2

Thus, the acceleration a when distance s=1 m is 25 m/s2 .

Given:

The distance is s=1 m .

The velocity equation is υ=(4t+5) m/s .

Write the velocity equation.

υ=(4t+5) m/s

Write the expression acceleration.

a=dυdt (I).

Here, acceleration is a and rate of change of velocity υ with respect to time (t) is dυdt .

Conclusion:

Substitute (4t+5) m/s for υ in Equation (I).

a=dυdt= ddt(4t+5) =4 m/s (II).

Thus, the acceleration a when distance t=2s is 4 m/s2 .

Given:

The time is t=2 s .

The acceleration is a=2 m/s2 .

The initial velocity is υ0=0

Write the expression for final velocity in y direction.

υ=υ0+act (III).

Here, final velocity is υ , initial velocity is υ0 , acceleration due to gravity is ac and time is t .

Conclusion:

Substitute 2 s for t , 0 for υ0 (s0) and 2 m/s2 for ac in Equation (I).

υ=υ0+act=0+(2)(2)=4 m/s

Thus, the velocity υ when time is t=2 s is 4 m/s .

Given:

The time is t=3 s .

The acceleration is a=2 m/s2 .

The initial velocity is υ0=3 m/s

The initial distance is (s0)=0 .

The final distance is s=4 m .

Write the expression for final velocity in y direction..

υ2=(υ02)+2ac(sy−(s0)) (II).

Here, final velocity is υ , initial velocity is (υ0)y , acceleration due to gravity is ac , final distance is s , initial distance is (s0)y .

Conclusion:

Substitute 3 m/s for υ0 , 4 m for s , 0 for (s0) and 2 m/s2 for ay in Equation (I).

υ2=(υ02)+2ac(s−(s0))υ2=(3)2+2(2)(4−0)υ=5 m/s

Thus, the velocity υ when distance is s=4 m is 5 m/s .

Given:

The distance is s1=5 m .

The distance is s2=4 m .

The acceleration equation is a=(s) m/s2 .

Write the acceleration equation.

a=(s) m/s2

Write the expression acceleration.

a=υdυdsads=υdυ (I).

Here, velocity is υ , acceleration is a and rate of change of velocity υ with respect to distance (s) is dυds .

Conclusion:

Substitute (s) m/s2 for a in Equation (I).

Integrate the Equation (I) at the limits 0 to υ for υ and s1=5 m to s2=4 m for s .

∫0υ dυ=∫45(s) ds[υ22]0υ=[s22]45

υ22=(5)22−(4)22υ2=25−16υ=3 m/s

Thus, the velocity υ is 3 m/s .

Given:

The time is t=3 s .

The acceleration is a=4 m/s2 .

The velocity is υ=2 m/s

The distance is (s0)=2 m

Write the expression for final distance in y direction.

s=(s0)+(υ)t+12a t2 (I).

Here, final distance is s , initial distance is (s0) , initial velocity is (υ) , acceleration due to gravity is a and time is t .

Conclusion:

Substitute 3 s for t , 2 m/s for υ , 2 m for (s0) and 4 m/s2 for ay in Equation (I).

s=(s0)+(υ)t+12a t2=2+2(3)+12(4) (3)2=26 m

Thus, the distance s when time is t=3 s is 26 m .

Given:

The time is t=1 s .

The acceleration equation is a=(8t2) m/s2 .

Write the acceleration equation.

a=(8t2) m/s2

Write the expression acceleration.

a=dυdta dt=dυ (I).

Here, acceleration is a and rate of change of velocity υ with respect to time (t) is dυdt .

Conclusion:

Substitute (8t2) m/s2 for a in Equation (I).

Integrate the Equation (I) at the limits 0 to υ for υ and 0 to t for t .

∫0υ dυ=∫0t(8t2) dt[υ]0υ=[8t33]0tυ=(2.667 t3) m/s (II).

Substitute 2 s for t in Equation (II).

υ=(2.667 t3) m/s=2.667(1)=2.67 m/s

Thus, the velocity when time is t=1s is 2.67 m/s .

Given:

The time is t=2s .

The distance equation is s=(3t2+2) m .

Write the distance equation.

s=(3t2+2) m

Here, average velocity is υavg , change in distance is Δs and change in time is Δt .

Write the expression velocity.

υ=dsdt (I).

Here, velocity is υ and rate of change of distance s with respect to time (t) is dsdt .

Conclusion:

Substitute (3t2+2) m for s in Equation (I).

υ=d dt(3t2+2)=6t m/s (II).

Substitute 2 s for t in Equation (II).

υ=6t m/s=6(2)=12 m/s 

Thus, the velocity υ when time is t=2s is 12 m/s .

Given:

The distance traveled by the particle from A to C is shown in Figure (1).

The time traveled by the particle from A to B is 4s .

The time traveled by the particle from B to C is 6s .

Write the expression for the average velocity.

υavg=ΔsΔt (I).

Here, average velocity is υavg , change in distance is Δs and change in time is Δt .

Write the expression for the average speed υsp .

υsp=sTotaltTotal (II).

Here, the total distance is sTotal and the total time traveled by the particle is tTotal .

Refer Figure (1) and calculate the total distance traveled by the particle.

sTotal=sAB+sBC (III)

Refer Figure (1) and calculate the total time traveled by the particle.

tTotal=tAB+tBC (IV)

Conclusion:

From the Figure (1) calculate the change in distance.

Δs=(s2−s1)=(6−(−1))=7 m

Calculate the change in distance

Δt=(t2−t1)=(10−0)=10 s

Substitute 10 s for Δt and 7 m for Δs in Equation (1).

υavg=ΔsΔt=7m 10s=0.7 m/s 

Thus, the average velocity of particle is 0.7 m/s .

The time traveled by the particle from A to B .

tAB=6 s

The time traveled by the particle from B to C .

tBC=4 s

Substitute 7 m for sAB and 14 m for sBC in Equation (III).

sTotal=sAB+sBC=7m +14 m=21m

Substitute 6 s for tAB and 4 s for tBC in Equation (IV).

ttotal=6 s+4 s=10 s

Substitute 21m for sTotal and 10 s for tTotal in Equation (II).

υsp=sTotaltTotal=21m10 s=2.1 m/s

Thus, the average speed of particle is 2.1 m/s