An ant starting at times = 40 cm zig-zags back and forth on a pienic table endin

Question

An ant starting at times = 40 cm zig-zags back and forth on a pienic table ending at times = 10 cm as shown. The ant's distance traveled and displacement is 50 cm and 50 cm. 30 cm and 50 cm. 50 cm and 30 cm. 50 cm and 50 cm. 50 cm and-30 cm. The slope at a point on a position-versus-time graph of an object is The object's speed at that point. The object's velocity at that point. The object's acceleration at that point. The distance traveled by the object to that point. I really have no idea. Here is a position graph of an object: At t = 1.5 s, the object's velocity is 40 m/s. 20 m/s. I0 m/s. -10 m/s. None of the above. A car starts from rest and reaches a speed of 20 m/s in 3.0 s (traveling in a straight line). Which of the following equations alone can be used to find the distance traveled Delta times = v_1t + 0.5at^2 v_t = v_1 + at v_t^2 = V_t^2 + 2a delta times delta times - 0.5(v_t + v_f)t There is NOT enough information. A ball is thrown straight up in the air. What is the acceleration of the ball when it reaches it maximum height 0 m -9.8 m/s^2 +9.80 m/s^2 There is NOT enough information

Explanation / Answer

1)

distance traveled = 20 + 10 + 20 = 50 cm

displacement = 40- 10 = 30 cm

So, option (c) is correct

2)

Slope of the position vs time curve gives the speed of the body

option (A) is correct

3)

velocity = slope at t =1.5 s

= 20/1

= 20 m/s

4)

distance is given by :

d = vi + 0.5*a*t^2

Now, a = (vf+vi)/t

So, d = 0.5((vi+vf)/t)*t^2

= 0.5*(vi+vf)*t <------- option d

5)

acceleration is constant = g = -9.8 m/s2

So, answer is b) -9.8 m/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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