Linear Motion -Velocity and Acceleration

[stryker23]

Linear Motion --Velocity and Acceleration

Homework Statement

[10] A toy car travels along a straight line. Its position x varies with time as shown on the x-t graph. The equation of motion in the A-C interval is given by

x(t)=alpha+(beta)t^2+y(t)^(1/2)

where alpha=1.0 m, beta=1.3 m/s^2 and y=0.3 m/s^(1/2) . The curve in the interval C-E is a symmetric parabola, and the graph in the E-G interval is linear.

a) Write the equations for v(t) and a(t)  in the A-C interval. Calculate the velocity in point C.
b) Calculate the accelerations in the intervals C-D and D-E, and the position in point D.

Homework Equations

x(t)=alpha+(beta)t^2+y(t)^(1/2)

There is a graph but I do not know how to show it.

The Attempt at a Solution

I really just want to make sure I am starting this correctly. For part (A) I filled in the given values and then took the first and second derivative to find the the velocity and the acceleration. Is that correct for this problem? The second part of a A asks for the velocity at C. "t" at "C" equals one so I took v(1). The answer I worked out was v(1)=(-2.6m/s^2)+(0.15m/s^(1/2)). This doesn't seem right to me. Could someone check to see if I am doing this the correct way? I do not know how to proceed at all with part B, so I would greatly appreciate a little direction.
Thank you.

 

[anathema]

Hello,

Yes, you are on the right track. To find the velocity and acceleration in the A-C interval, you can use the equations:

v(t) = dx/dt = 2beta*t + (1/2)y^(1/2)

a(t) = dv/dt = 2beta + (1/4)y^(-1/2)*dy/dt

Substituting in the given values, we get:

v(t) = -2.6t + 0.15 m/s^(1/2)

a(t) = -2.6 m/s^2 + (0.75/y^(1/2)) m/s^3

To find the velocity at point C, you can substitute t=1 into the equation for v(t), which gives you v(1) = -2.45 m/s.

For part B, you can use the equation for acceleration to find the accelerations in the C-D and D-E intervals. For the position at point D, you can use the equation for position and substitute t=2.5 to find x(2.5).

I hope this helps. Let me know if you have any further questions.

 

[Moetasim]

Yes, you are on the right track in solving for the velocity and acceleration in the A-C interval. However, there seems to be a mistake in your calculation for the velocity at point C. The correct equation for velocity in this interval is v(t) = 2(beta)t + 0.5y(t)^(-1/2). Plugging in the given values, we get v(1) = -2.6 m/s.

For part B, you can use the equation a(t) = 2(beta) to calculate the acceleration in the C-D and D-E intervals. In the C-D interval, t = 2, so a(2) = -2.6 m/s^2. In the D-E interval, t = 3, so a(3) = -2.6 m/s^2.

To find the position at point D, you can use the equation x(t) = 1 + (beta)t^2 + y(t)^(-1/2). Plugging in t = 2, we get x(2) = 1.3 m.

I hope this helps! Keep up the good work.