If the average velocity of an object is zero in some time interval, what can you

Question

If the average velocity of an object is zero in some time interval, what can you say about its displacement in that interval? what about its distance travelled?

Problem 1. If the average velocity of an object is zero in some time interval, what can you say about its displacement in that interval? What about its distance travelled? Explain your answers Problem 2. Is it possible for a particle to have a velocity and an acceleration in opposite directions? If so, sketch a velocity-time graph proving your point. If not, explain why not. Problem 3. A runner runs in a straight line with an average velocity of 5.t m/s for 5 minutes and then with an average velocity of 4x m/s for 4 minutes. (a) What is her total displacement? (b) What is her average velocity during this time? (c) Suppose her goal is to run 5 km in 15 minutes. hat average velocity must she have over the remaining portion of her run? Problem 4. A ball is thrown upward. While the ball is in the air, (a) does its acceleration increase, decrease or remain the same? Explain your answer. (b) Describe what happens to its velocity as accurately as you can Problem 5. The position of a pinewood derby car was observed at various times as shown in the table below. Find the average velocity of the car for (a) the first second, (b) the last 3 seconds and (c) the entire period of observation r (m) 0.0 2.3 9.2 20.7 36.8 57.5 t (s 0.0 1.0 2.0 3.04.0 5.0

Explanation / Answer

Problem 6) Given that,

x = 2t + 3t2

We need to calculate the instantaneous velocity and acceleration at t = 3 sec.

We know that, v = dx/dt

v = dx/dt = d( 2t + 3t2)/dt = 2 + 6t

v = 2 + 6 x 3 = 20 m/s

We know that, a = dv/dt

a = d(2 + 6t )/dt = 6 m/s2

Hence, v = 20 m/s and a = 6 m/s2

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Problem 13) Given that,

vi = 80 m/s ; a = 4 m/s2 ; H = 1000 m

A) From equation of motion we know that,

Y = ut + 1/2 gt2

In our case, Y = 1000 ; u = vi = 80 m/s and a = 4 ; t is to be determined. Putting values we get

1000 = 80 t + 1/2 x 4 xt2 => 2 t2 + 80 t - 1000 = 0

The above quadratic equation gives us: t = 10 and (-50 is neglected as t is not negative)

In the following parts we got to find time at diffrent stages of its travel,

So the total time of travel will be = 10 + 12 + 18.8 = 40.8 sec

we know that, vf = vi + at => vf = 80 + 4 x 10 = 120 m/s

(b)Let H be the max height reached.

Now its falling freely, so a = -g = -9.8 m/s2. and its velocity is vf = 120 m/s

again by vf = vi + at => t =( vf - vi) / a = (120 - 0 ) / 9.8 =12 sec

again using,H = Yo + vft + 1/2 gt2

H= 1000 + 120 x 12 - 4.9 x 144 = 1734 m

Hence, H = 1734 meters

(c)Let V be the velocity.

H = Yo + vft + 1/2 gt2 At ground H = 0 ; Yo = 1734 and t we need to find v = 0 as its velocity will be zero at the higest point.

0 = 1734 + 0 - 1/2 x 9.8 x t2 => t = 18.8 sec

Again by using, v = u + at

V = 0 + 9.8 x 18.8 = 184.24 m/s

Hence, V = 184.24 m/s

[PS: There are too many questions here, I have done 2 for you, Pls put the others seperately... thanks].

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