physics problem

I'am very glad because I can ask you melody and alan you always help me I'm very Grateful 

 

so in addition to the mistake which Alan mentioned it also I was thinking that ' t ' is the last second of motion 

so I said 10 s .. but now you told me that ' t ' is the period which 5 s ... 

 

 

thank you 

At the beginning of the second 5 second period the velocity is 5a    (so u=5a)

We are looking at the next 5 seconds so t=5

the distance travelled during this 5 seconds is 150 m

 

Using the formula

 

$$\\s=ut+\frac{1}{2}at^2\\\\
150=5a\times 5+\frac{1}{2}a\times5^2\\\\
150=25a+\frac{25a}{2}\\\\
300=50a+25a\\\\
300=75a\\\\
a=4\\\\$$

I've gotten a bit carried away here - have I help you with your dilemma?

xvxvxv wrote: how when x-x0 = 150   , t=5s ?  I think t must equal to 10 s ? 

 

The total time from the start when the ball was at rest is indeed 10 seconds.  However, the problem is solved in such a way that time has been reset to zero for the second part of the problem.  I can understand how you might have been misled, because the question doesn't reset the start of the distance to zero - it calls the start of the second part x₀.  If you like you could solve the question using t set at 10 seconds, but then you must use the equation in the form:  x-x₀ = v_0x(t-t₀)+(1/2)a_x(t-t₀)², where t₀ is 5 seconds.

 

.