The curvilinear motion of a particle is defined by v, = 50 – 16t and y =100 – 4t’, where v, is in meters per second, y is in meters, and t is in seconds.It is also known that x = 0 when t – 0. Plot the path of the particle and determine its velocity and acceleration when the position y = 0 is reached.

To plot the path of the particle, we need to find the expression for x in terms of t. We know that the velocity v_x is given by:

vx=dxdt=50−16t𝑣𝑥=𝑑𝑥𝑑𝑡=50−16𝑡

Integrating with respect to t, we get:

∫dx=∫(50−16t)dt∫𝑑𝑥=∫(50−16𝑡)𝑑𝑡

x=50t−8t2+C𝑥=50𝑡−8𝑡2+𝐶

Given that x = 0 when t = 0, we can find the value of the constant C:

0=50(0)−8(0)2+C0=50(0)−8(0)2+𝐶

C=0𝐶=0

Therefore, the expression for x in terms of t is:

x=50t−8t2𝑥=50𝑡−8𝑡2

Now, to determine the velocity and acceleration when y = 0, we need to find the time t when y = 0. Given that y = 100 – 4t, we can set y = 0 and solve for t:

0=100−4t0=100−4𝑡

4t=1004𝑡=100

t=25𝑡=25

At t = 25, y = 0. Now, we can find the corresponding x value by substituting t = 25 into the expression for x:

x=50(25)−8(25)2𝑥=50(25)−8(25)2

x=1250−5000𝑥=1250−5000

x=−3750𝑥=−3750

Therefore, when y = 0, the particle is at the point −3750,0−3750,0.

To find the velocity at this point, we can substitute t = 25 into the expression for v_x:

vx=50−16(25)𝑣𝑥=50−16(25)

vx=50−400𝑣𝑥=50−400

vx=−350𝑣𝑥=−350

So, the velocity of the particle when y = 0 is -350 m/s.

To find the acceleration at this point, we differentiate the expression for v_x with respect to t:

ax=dvxdt=−16𝑎𝑥=𝑑𝑣𝑥𝑑𝑡=−16

Therefore, the acceleration of the particle when y = 0 is -16 m/s².