7. The x-coordinate of a particle in curvilinear motion is given by where x is in
meter and t is in seconds. The y-component of acceleration in meter per second squared is
given by If the particle has y-components and when
find the magnitudes of the velocity v and acceleration a when t=0.2*041 . sec
8. The y-coordinate of a particle in curvilinear motion is given by where y is
in centimeter and t is in seconds. Also, the particle has an acceleration in the x-direction
given by cm/sec2. If the velocity of the particle in the x-direction is 1.4*041cm/sec
when t=0, calculate the magnitudes of the velocity v and acceleration a of the particle when
Construct v and a in your solution. t = 1 sec.
7. To find the magnitudes of velocity v and acceleration a at t = 0.2*041 seconds, we need to differentiate the x-coordinate equation with respect to time t to find the velocity components vx and vy, and then differentiate the velocity components with respect to time to find the acceleration components ax and ay.
Given x = (t^2 - 0.2t) meters,
Differentiating x with respect to t:
vx = d(x)/dt = 2t - 0.2
Given y = 0.4t^2 - 0.06t^3 meters,
Differentiating y with respect to t:
vy = d(y)/dt = 0.8t - 0.18t^2
Given ay = 1.5 meters per second squared,
Given vy = 2 meters per second,
We can find the acceleration components at t = 0.2*041 seconds:
ay = 1.5 m/s^2
vx = 2 m/s
vy = 0.8(0.2*041) - 0.18(0.2*041)^2 = 0.0824 - 0.002259 = 0.080141 m/s
The magnitude of velocity v is:
v = sqrt(vx^2 + vy^2) = sqrt((2^2) + (0.080141^2)) = sqrt(4 + 0.006426) = sqrt(4.006426) = 2.0016 m/s
The magnitude of acceleration a is given by:
a = sqrt(ax^2 + ay^2) = sqrt((0^2) + (1.5^2)) = sqrt(0 + 2.25) = sqrt(2.25) = 1.5 m/s^2
Therefore, at t = 0.2*041 seconds, the magnitude of velocity v is 2.0016 m/s and the magnitude of acceleration a is 1.5 m/s^2.
8. Given y = 3t^2 - 0.06t^3 centimeters and ax = 0.8 cm/sec^2, we first need to differentiate y with respect to t to find the velocity component vy, and then find the velocity components vx and vy.
Given y = 3t^2 - 0.06t^3 cm,
Differentiating y with respect to t:
vy = d(y)/dt = 6t - 0.18t^2
Given vx = 1.4*041 cm/sec,
We can find the velocity components at t = 1 second:
vx = 1.4*041 cm/sec = 1.4*041 cm/s
The magnitude of velocity v at t = 1 second is:
v = sqrt(vx^2 + vy^2) = sqrt((1.4*041)^2 + (6 - 0.18)^2) = sqrt(1.96 + 36 - 0.36 +) = sqrt(37.6) ≈ 6.14 cm/s
To find the acceleration vector a, we differentiate the velocity components with respect to time:
ax = d(vx)/dt = d(1.4*041)/dt = 0
ay = d(vy)/dt = d(6t - 0.18t^2)/dt = 6 - 0.36t
The magnitude of acceleration a at t = 1 second is:
a = sqrt(ax^2 + ay^2) = sqrt(0 + (6 - 0.36*1)^2) = sqrt(0 + 5.64) ≈ 2.38 cm/s^2
Therefore, at t = 1 second, the magnitude of velocity v is approximately 6.14 cm/s and the magnitude of acceleration a is approximately 2.38 cm/s^2.