Average Acceleration without given Velocities and displacements?

Question

In 5 seconds, a dog accelerates to its max speed, then maintains that speed for 10 seconds to reach his trainer, who was 50 meters away from the starting point of the dog. What was it acceleration for the first 5 second? For the whole 15 seconds?

Answer 1

a - acceleration
s - distance after 5 seconds
v - speed after 5 seconds

Now we have to solve equations:

v = 5 a
s = (25 a)/2
50 - s = 10 v

Do it, it's easy.

Answer 2

Here's my plan:

To get the acceleration for the first five seconds, we'll utilize three fundamental equations, namely:
s = (1/2)*a*t^2
s = v*t
v = a * t
(Note: Those equations only work when the acceleration is constant.)

To calculate the acceleration for the whole 15 seconds, we'll take a weighted average of the acceleration for the first 5 seconds and for the last 10 seconds (which happens to be zero).

So, let's calculate the acceleration for the first 5 seconds:
Let s1 = the distance traveled during the first 5 seconds. Then we have s1 = (1/2)*a*t^2 = (1/2) * a * 5^2 = (25/2)*a.
So, s1 = (25/2) * a

To solve for a (the acceleration during the first five seconds), we need to know what s1 is.
So, we have distance = velocity * time, namely (50 - s1) = v * 10 for the second part of the trip.
But v (i.e. the max velocity attained) is just the acceleration during the first five seconds times the time of that acceleration. Therefore
(50 - s1) = (a * 5) * 10
(50 - s1) = 50 * a
s1 = 50 - 50a
Substituting into s1 = (25/2) * a, we have
50 - 50a = (25/2) * a
100 - 100a = 25 * a
100 = 125 * a
a = 100/125 = 4/5

Therefore the acceleration for the first 5 seconds is 4/5 m/s^2.

The acceleration for the whole 15 seconds is:
(4/5) * (5/15) + 0 * (10/15)
= 4/5 * (1/3) + 0 * (2/3)
= 4/15 m/s^2

I hope this helps.

Answer 3

It's max speed is 50m / 10s = 5m/s.

Thus, since the dog accelerated to 5m/s in 5 seconds (it started at rest), its average acceleration during the first 5 seconds was 5m/s / 5s = 1m/s^2.

To find the average acceleration for the entire 15 seconds, we need change in speed / change in time. Thus, the dog's average acceleration over the entire 15 seconds would be 5m/s / 15s = 1/3 m/s^2.

Answer 4

the statement is correct and if you know vector and scalar definitions then it is easy a quantity with both magnitude and direction is called a vector but a quantity with only magnitude and no direction are scalars see, speed while you calculate speed you will notice that you calculate total distance /total time which is just magnitude but while calculating velocity in you thing i.e, a to b to c to a, you tell that displacement and hence velocity is zero because you see the direction of motion of the particle ( after you read vectors you will know about them) to consider that velocity is zero otherwise you will tell that velocity is just equal to total distance /total time which gives speed as per the definitions of vectors and scalars, you see that vectors are an addition of scalars and magnitude (not fully) i.e, velocity which is a magnitude + direction but speed is just magnitude so, you can say that velocity is speed with direction only in a straight line motion but in other type of motions also the statement is correct as per the definitions of vectors and scalars so, in your question, velocity =0 also it is not a straight line motion but the statement as told is right just because velocity and speed both have the same formula and just differ in direction which if given to speed it becomes velocity just for telling. also displacement, velocity , and accelerations are vectors acceleration is = rate of change of velocity /time taken it means the amount by which velocity has changed in a given time interval so, it is also a vector acceleration = dv/dt i.e, acc. = differentiation of velocity with time and velocity =ds/dt where s=displacement in your question, displacement =0 i.e, change of position or displacement =0 so, rate of change of displacement is also 0 hope you understand