An experiment done on the Moon shows the position of a particle moving along y-a

Question

An experiment done on the Moon shows the position of a particle moving along y-axis varies in time according to the expression Y(t)= 15 + 6t - 8t^2, where Y is in cm and t is in seconds. What are the initial velocity and initial position of the particle? Write an expression and draw a graph of the velocity as a function of time? Determine the acceleration, velocity and position of the particle at t = 3 s. In a free fall experiment, a ball is thrown downwards with initial velocity of 30 m/s. Determine the velocity and distance traveled by the ball after 4, 15, 30 and 50 seconds. Compute the maximum height reached by the ball. A car travels 30 km due north and then 40 km due west. Find the magnitude and direction of the car's resultant displacement? Given the vectors N = 3 i + 4 j and M = 3i-4 j. Determine L = N+M. Determine F = N - M. Find the magnitude and direction of the vectors F and L. Draw them. Three displacement vectors of a croquet ball are shown in figure 1 where A=6 m, B=10 m and C=8 m. Find (a) the resultant in unit-vector notation, and (b) the magnitude and direction of the resultant displacement. It took 4 s to come to stop for a car to travel 20 m. Find its initial velocity and acceleration. Three displacement are D = 3.0 due north, E = 4 m due east and G = 8 due east of south. Draw a diagram for R = D + E + G. What are the magnitude and direction of vector R?

Explanation / Answer

ANSWER

Keep in mind that because of terms and conditions of chegg.com we, the tutors, can not answer questions with more than four (4) sub-questions. This problem have seven (7) questions, I will answer only four questions but I recommend to you that upload other question to the system with the other exercises, and another tutor will answer it.

Problem 2.

at 4 s,

v= 30-4*9.8 = -9.2 m/s
s = 30*4 -4.9*4^2 = 41.6 m

at 15 s,

v= 30-15*9.8 = -117 m/s
s = 30*15 -4.9*15^2 = -652.5 m

at 30 s,

v= 30-30*9.8 = -264 m/s
s = 30*30 -4.9*30^2 = -3510 m

at 50 s,

v= 30-50*9.8 = -460 m/s
s = 30*50 -4.9*50^2 = -10750 m

Problem 3.

the magnitude of resultant displacemnt (let x ) is

x^2 = 30^2 + 40^2

x = 50 km

and direction will be

tan theta = 30/40 = 3/4

theta that is the angle will be

theta = tan^-1 3/4 = 36.87°

Problem 4.

(a)

L = (3i+4j) + (3i-4j) = (6i+0j)

(b)

F = (3i+4j) - (3i-4j) = (0i+8j)

(c)

L (Magnitude) = sqrt(62+02) = 6

L (Direction) = on x-axis (to the right)

F (Magnitude) = sqrt(02+82) = 8

F (Direction) = on y-axis (upwards)

Problem 6.

(v+u)t/2 = s

(0+u)4/2 = 20

u = 10 m/s

a = (0 - 10)/4 = -2.5 m/s2

Regards!!!

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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