An object starts from rest at the origin and has an acceleration equal to 10 m/s
ID: 1652841 • Letter: A
Question
An object starts from rest at the origin and has an acceleration equal to 10 m/s in the y direction and 5 m/s in the x direction. After 20 s:
What is its velocity in x and y components?
A. 100 m/s in x and 200 m/s in y
B. 200 m/s in x and 100 m/s in y
C. 400 m/s in x and 50 m/s in y
D. 800 m/s in x and 10 m/s in y
What is its position in x and y components?
A. 1000 m in x and 2000 m in y
B. 2000 m in x and 1000 m in y
C. 500 m/s in x and 1000 m/s in y
D. 1000 m/s in x and 500 m/s in y
What is its velocity in magnitude and direction?
A. 150 m/s @ 40 degrees from the x-axis
B. 224 m/s @ 63 degrees from the x-axis
C. 300 m/s @ 75 degrees from the x-axis
D. 500 m/s @ 30 degrees from the x-axis
What is its displacement in magnitude and direction?
A. 1500 m @ 40 degrees from the x-axis
B. 2240 m @ 63 degrees from the x-axis
C. 3000 m @ 75 degrees from the x-axis
D. 5000 m @ 30 degrees from the x-axis
Explanation / Answer
1) velocity in X is Vx = Vo+(ax*t)
given that initial velocity Vo = 0 m/sec
ax = 10 m/s^2
t = 20 sec
then
Vx = 0+(10*20) = 200 m/sec
velocity in y is Vy = Vo+(ay*t) = 0 +(5*20) = 100 m/sec
so the answer is B.200 m/sec in x and 100 m/s in y
2) position in x is
x = (Vo*t) + (0.5*ax*t^2)
x = (0*t)+(0.5*10*20^2) = 2000 m
position in y is
y = (Vo*t)+(0.5*ay*t^2)
y = (0*t)+(0.5*5*20^2)
y = 1000 m
so the answer is B. 2000 m in X and 1000 m in Y
velocity in magnitude is V = sqrt(Vx^2+Vy^2) = sqrt(200^2+100^2) = 224 m/sec
direction is theta = tan^(-1)(100/200) = 27 degrees
from X-axis is 90-27 = 63 degrees
so the answer is B.224 m/s @63 degrees from the X-axis
dispalcement is s = sqrt(x^2+y^2) = sqrt(2000^2+1000^2) = 2240 m
direction is theta = tan^(-1)(1000/2000) = 27 degrees
but from the X-axis is 90-27 = 63 degrees
so the answer is B
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