Answer:
Part a)


Part b)

Explanation:
Part a)
Constant speed by which the student will run is given as

now after some time if student is going to overtake the position of bus
so here the final positions will be same
so we have




so it is

So student will run the total distance



Part b)
Speed of bus when student reach the bus is given as



Answer:
25 m/s
Explanation:
First we should define the variables
T=4
Dx = 100
ay=-9.8
ax=0
We can use formula 1 from the BIG 5
x=(v+v0)t/2
By plugging in our variables we can get 100=4(v+v0)/2
Which is 50=v+v0
v=v0 since horizontal acceleration always equals zero
so 2v0 = 50
v0 = 25
You can use photo math for This
Answer:
The time where the avergae speed equals the instaneous speed is T/2
Explanation:
The velocity of the car is:
v(t) = v0 + at
Where v0 is the initial speed and a is the constant acceleration.
Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:

v_ave = v0+a(T/2)
We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>
v(T/2)=v_ave
Now we want to know where the car should be, the osition of the car is:

Where x_A is the position of point A. Therefore, the car will be at:
<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>