Just number 1 will do. Im not exactly sure as to what to do and my teacher doesn

Question

Just number 1 will do. Im not exactly sure as to what to do and my teacher doesnt explain it very well. If someone could write out step by step as to what to do that would be great. I dont just want the answer.

A ball moves in a straight line along an x-axis with a constant acceleration of 3.50 m/s2. After 2.00 s, it has attained a speed of 14.5 m/s. Determine the displacement of the ball between times 2.00s and 5.00 s. Determine the total time it takes the ball to move a distance of 50.0 m from its start. Determine the velocity of the ball when it moved a distance of 50.0 m from its start. The position of a particle moving in one-dimension is described by the following formula: x = 12.0t3 - 3.00t2 + 14.0 (x is in meters, t is in seconds, so velocity is in m/s and acceleration is in m/s2) Complete the entries in the table below for the position, instantaneous

Explanation / Answer

we have

v= u + a t

14.5= u + 3.5x2

u=7.5 m/s (that is its initail velocity )

-----------------------------------------================================---------------------------------------------

(A) For this question we use the formula

S=u t + 1/2 a t^2

where we have

u= 14.5 m/s

t=5-2=3 sec

a=3.5 m/s^2

so we have displacement

S=14.5 x3 +1/2 x3.5x3^2=59.25 m

---------------------------------------------------============================--------------------------------------

(b) S=ut + 1/ a t^2

S=50

u=7.5

a=3.5

so

50=7.5 t +.5*3.5 t^2

now solve for t in this quadratic equation

------------------------------------------------===================================---------------------------------------------

v^2-u^2=2aS

so we have v^2=2x3.5*50-7.5^2

v=sqrt(293.75)

v=17.139 m/s

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.