Question

.A particle moves in the X –Y plane so that its x coordinate is defined by
x = 5 t3 – 105 t where x is in cm and t is in seconds. When t = 2 s the total acceleration is
7*0.41m/s2. If the y component of acceleration is constant and the particles starts from rest at
the origin when t = 0, determine its velocity when t = 4 sec.

Answer 1

s_x = 15t^3 105t
v_x = 45t^2 - 105
a_x = 90t

at t=2, a_x(2) = 180
a = √(a_x^2 + a_y^2) = √(180^2 + a_y^2) = 2.87
how can this be, since a_y^2 is positive?

something's wrong with the problem as presented.

Answer 2

To find the velocity when t = 4 seconds, we need to find the x and y components of the velocity at that time.

Given:
x = 5t^3 - 105t
dx/dt = 15t^2 - 105

At t = 4 seconds:
dx/dt = 15(4)^2 - 105
dx/dt = 15(16) - 105
dx/dt = 240 - 105
dx/dt = 135 cm/s

Now, we need to find the y component of acceleration:
Total acceleration, a = 7(0.41)
a = 2.87 m/s^2

Since the particle starts from rest and the acceleration is constant, the y component of acceleration is:
a_y = a
a_y = 2.87 m/s^2

Now, we can find the y component of velocity at t = 4 seconds:
a_y = dv_y/dt
dv_y = a_y * dt
v_y = a_y * t
v_y = 2.87 * 4
v_y = 11.48 m/s

Therefore, the velocity of the particle at t = 4 seconds is:
Velocity = sqrt((dx/dt)^2 + v_y^2)
Velocity = sqrt((135)^2 + (11.48)^2)
Velocity = sqrt(18225 + 131.7504)
Velocity = sqrt(18356.7504)
Velocity ≈ 135.51 cm/s

So, the velocity of the particle when t = 4 seconds is approximately 135.51 cm/s.

Answer 3

You are correct, there seems to be a mistake in the problem as presented. Let's try to correct it.

Given:
x = 15t^3 - 105t
v_x = 45t^2 - 105
a_x = 90t

At t=2 seconds:
a_x = 90(2) = 180 cm/s^2

The total acceleration is defined as:
a = √(a_x^2 + a_y^2)

Given that a = 2.87 m/s^2 (287 cm/s^2), we can rewrite the equation as:
287^2 = 180^2 + a_y^2
82369 = 32400 + a_y^2
a_y^2 = 49969

This gives us the acceleration in the y-direction (a_y) as the square root of 49969 which is approximately 223.57 cm/s^2.

Now, let's find the velocity when t=4 seconds:
v_x = 45(4)^2 - 105
v_x = 45(16) - 105
v_x = 720 - 105
v_x = 615 cm/s

v_y = a_y*t
v_y = 223.57 * 4
v_y = 894.28 cm/s

The total velocity is then:
v = √(v_x^2 + v_y^2)
v = √(615^2 + 894.28^2)
v = √(378225 + 800726.46)
v ≈ √(1179951.46)
v ≈ 1086.87 cm/s

Therefore, the velocity of the particle when t = 4 seconds is approximately 1086.87 cm/s. Thank you for pointing out the mistake earlier.