# Instantaneous and Average speeds and Average acceleration.

A runner is jogging at a steady velocity of 3.3 km/hr. When the runner is at 8.6Km from the finishing line, a bird begins flying from the runner to the finishing line at 13.2 km/hr. When the bird reaches the finish line, it returns around and flies back to the runner. How far does the bird travel? Answer in units of Km..... and 12 other problems.

© BrainMass Inc. brainmass.com October 25, 2018, 1:38 am ad1c9bdddfhttps://brainmass.com/physics/acceleration/instantaneous-average-speeds-average-acceleration-269730

#### Solution Preview

Step 1 : Now the time taken by the bird to reach the finishing line will be

t1 = distance / velocity = 8.6 km / 13.2 kms-1 = 0.6515 sec.

But the person is not at rest. So he will travel some distance in the same time.

Distance traveled by the person in the time of 0.6515 sec is

D1 = 3.3 * 0.6515 = 2.15 km.

Hence remaining distance will be

d = 8.6 - 2.15 = 6.45 km .

STEP 2: Now the person and the bird are at quite opposite directions. As shown in the figure.

Suppose that they reach ...

#### Solution Summary

Average speed or velocity is a measure of overall fastness of motion during a specified interval of time.

The average velocity of a particlefor a given interval of time is defined as the ratio of displacement traveled to the time taken.

Average Velocity = Displacement / Time.

It is required to discuss some special cases here.

Case i) : If the particle travels distances L1,L2 etc at speeds v1,v2,.. etc., respectively, then, as

Net displacement

s = L1+L2+ L3+........+Ln = Ln

And time taken will be

.

Special case i) If the distances traveled are the same (i.e when L1=L2=....=L), then the above equation reduces to

Special case ii) If n = 2 average velocity cab calculate from

Case ii) If the particle travels at speeds v1,v2, etc for intervals t1,t2,.. etc then as

s = v1t1 + v2t2+.....+vntn

And t = t1+t2+.....+tn

Then

Special case iii) If t1 = t2 = ..... = tn = t then the above equation reduces to

Now here are the answers to your questions.

Q1) Here car is driven for the first 7 hrs with 46 km/h and next 7 hrs with 69 km/h. Hence case ii will be suitable to this. According to this,

=

Q2 ) Here first 402.4 km is traveled with 46 kmph and next same distance with a speed of 69 kmph. Here the first case will be suitable for the problem.

According to the special case ii), when distances are equal we can use, .

Here solving we get,

Vavr = = 55.2 km/h

Q3) In this question imagine that the lead car is 'L' and the car we are talking about is 'C'.

Now the car C has to complete 52 laps as it is 1 lap behind the L. And the total distance it has to travel is

d = 52 * ( dist of one lap ) = 52 * 1.72 = 89.44 km. = 89440 m.

Here in the problem the condition is to bring both the cars to the finishing line at the same time. So find the time taken by the leading car to finish the 51 laps ( ie a distance of 51*1720 m = 87720 m) with a speed of 53.6 ms-1, will be

Time = dist / velocity = 87720 / 53.6 = 1636.57 seconds.

Now the car C should complete its journey (ie., 52 laps = 89440 m) in the same time.

Now average speed required by the second car C is

Q4) Even this question is almost same as the question No: 3. Here from the same source ( earth quake ), the both of the waves started. But one wave (longitudinal) moves faster, with 7.2688 kms-1 than the transverse wave ( 4.13 kms-1 )

Here the dist traveled by both of them will be the same. If the T wave spends a time 't' to travel a distance 'd' the L-wave takes 83 seconds lesser to travel the same distance 'd' as it is faster.

So dist (d) = (7.2688)(t - 83) = 4.13(t)

 (7.2688-4.13)(t) = 7.2688 * 83

Solving we get t = 192.21 sec

Here d = 4.13 kms-1 * 192.21 sec = 793.83 km.

This is the distance at which the earth quake has developed.

Q5) Here the reconnaissance plane traveling some unknown dist ( to and fro ) with two different speeds. But as the dist is same we can get the average speed using .

So ms-1

Q6) It is a good question but a bit lengthy. It is given that the dist between the person and the finishing line is 8.6 km and speeds of the person and the bird are 3.3 km/hr and the other is 13.2 km/hr ( 4 times 3.3 km/hr ).

Step 1 : Now the time taken by the bird to reach the finishing line will be

t1 = distance / velocity = 8.6 km / 13.2 kms-1 = 0.6515 sec.

But the person is not at rest. So he will travel some distance in the same time.

Distance traveled by the person in the time of 0.6515 sec is

D1 = 3.3 * 0.6515 = 2.15 km.

Hence remaining distance will be

d = 8.6 - 2.15 = 6.45 km .

STEP 2: Now the person and the bird are at quite opposite directions. As shown in the figure.

Suppose that they reach each other at a distance of x from the person. In a time t. Hence the following equation arises.

X = 3.3 * t

And 6.45 - x = ( 13.2 *t )

Soving the above equations we get t = 6.45 / 16.5 = 0.39 sec.

Hence the distance traveled by the bird will be 6.45 - ( 3.3 * 0.39)

= 5.16 km

Here bird in the forward journey reached the finishing line once and then came back by the distance of 5.16 km.

Hence the total distance traveled by the bird will be 13.76 km

Q7) When two of them are moving it is better to imagine man at rest and take the relative velocity of the bird relative to the man.

This will be

( 13.2 - 3.3) = 9.9 k.ms-1 while they are moving in same direction and

The relative velocity will be

( 13.2 + 3.3) = 16.5 k.ms-1 while they are moving in opposite direction.

When only the person is allowed to move, he will take a time of

t = 8.6 km / 3.3 kms-1 = 2.606 sec.

Now if the person is imagined at rest, the distance traveled by the bird in this interval will become as the answer to our question.

So distance traveled by the bird is d = 13.2 * 2.606 sec = 34.4 km.

Q8) This is an easy question. Notice in the question that the distances are divided into equal halves. Hence our formula used in special case i ( ) can be used . Here the total distance is divided in to two halves. Hence n = 2 . Hence the eqn becomes as

Since as given vavr = 91 kmph and v1 = 57 kmph. Then we need to calculate the v2.

Solving we get as

Q9) In the electric field, the electrons get a uniform force. Hence a uniform acceleration which can be calculated from the eqn

In our question u = the initial velocity = 105 ms-1 ,

v = the final velocity = 1.5 x 108 ms-1

s = the displacement = 1.8 cm = 1.8 x 10-2 m

By applying these in the above formula we have

as the answer.

Q10) Now we need to calculate the time taken in the above process.

For this it will be useful if we consider as displacement = average velocity * time

But when the acceleration is constant, there is a formula according to which we can write as

On solving we get t as

There fore t = 24 x 10-11 seconds.

STEP 1 : Now imagine that the person stays at the starting line and the bird only moves with a velocity of 9.9 kms-1 to reach the finishing line, it takes a time given by

t1 = dist / velocity = 8.6 km / 9.9 kms-1 = 0.8686 sec.

But the person is not at rest. So he will travel some distance in the same time.

Distance traveled by the person in the time of 0.8686 sec is

D1 = 3.3 * 0.8686 = 2.867 km.

Hence remaining distance will be

d = 8.6 - 2.867 = 5.733 km .

Physics: Density, Vectors, Speed, Acceleration, Velocity

See attached file for proper formatting.

1.7. Which of the following equations are dimensionally correct? (a) vf = vi + ax

(b) y = (2 m)cos(kx), where k = 2 m-1.

1.26. The radius of a solid sphere is measured to be (6.50 ± 0.20) cm, and its mass is measured to be (1.85 ± 0.02) kg. Determine the density of the sphere in kilograms per cubic meter and the uncertainty in the density.

1. 44. Vector has x and y components of -8.70 cm and 15.0 cm, respectively; vector has x and y components of 13.2 cm and -6.60 cm, respectively. If, what are the components of ?

2. 12. An object moves along the x axis according to the equation x(t) = (3.00t2 - 2.00t + 3.00) m, where t is in seconds. Determine (a) the average speed between t = 2.00 s and t = 3.00 s, (b) the instantaneous speed at t = 2.00 s and at t = 3.00 s, (c) the average acceleration between t = 2.00 s and t = 3.00 s, and (d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s.

2. 25. A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck travels for 20.0 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 5.00 s. (a) How long is the truck in motion? (b) What is the average velocity of the truck for the motion described?

2. 55. A rock is dropped from rest into a well. The sound of the splash is heard 2.40 s after the rock is released from rest. How far below the top of the well is the surface of the water? The speed of sound in air (at the ambient temperature) is 336 m/s.

View Full Posting Details