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Question # 12.
A particle is moving along a straight line such that its position is defined by s = (10t^2 + 20) mm, where t is in seconds. Determine (a) the displacement of the particle during the time interval from t = 1 s to t = 5 s, (b) the average velocity of the particle during this time interval, and (c) the acceleration when t = 1 s.
Solution Question # 12.
The average velocity from t = 1 s to t = 5 s is Vavg = △s / △ t = [s (5) - s (1)] / [ (5) - (1)] = [{10(5)^2 +20} – {10(1)^2 +20}] / 4 = 60 mm/s Differentiate the given position function to determine the velocity. V = ds / dt V = d (10t^2 + 20) / dt V = 20t Integrate the velocity from t = 1 s to t = 5 s to find the displacement during this interval. △s = (^) ∫ 1 5 v (^ t )^ dt △s = (^) ∫ 1 5 ( 20 t ) dt △s = 10t^2 |^51 △s = 10(5^2 - 1^2 ) △ s = 240 mm Differentiate the velocity to determine the acceleration. a = dv/dt a = d (20t) / dt a = 20
Therefore, the acceleration of the particle at t = 1 s is a (1) = 20 mm/s^2
Question # 12.
A particle moving along a straight line is subjected to a deceleration a = (-2v^3 ) m/s^2 , where v is in m/s. If it has a velocity v = 8 m/s and a position s = 10 m when t = 0, determine its velocity and position when t = 4 s.
Solution Question # 12.
The acceleration of the particle is written as: a = dv / dt -2v^3 = dv / dt (- 1 / 2v^3 ) dv = dt ∫ dt^ =
∫ dv v 3 ¿
t = (1/4v^2 ) + C Putting t = 0 and v = 8 m/s 0 = 1/4*8^2 + C C = -3.906 x 10- Now, t = (1/4v^2 ) + C 4 = (1/4v^2 ) + -3.906 x 10- v = 0.25 m/s Putting C = 0 in t = (1/4v^2 ) + C, v becomes v = 1 / sqrt (4t)
Solution Question # 12.
As the acceleration is not constant, we will need to integrate our acceleration to figure out the velocity. To do so, we will need to use the following equation: ads = vdv Take the integral of both sides: ∫ s 0 s ads =∫ v 0 v v dv ∫ 0 s ( 6 +0.02 s )=∫ v 0 v vdv (6s + 0.02s^2 / 2) |s 0 = (v^2 /2) |v 0 (6s + 0.02s^2 / 2) = (v^2 /2) v = sqrt (12s + 0.02s^2 ) When the height is 2000 m, the velocity is:
v = sqrt [12(2000) +0.02(2000)^2 ] v = 322.5 m/s To find the time, remember that: v = ds/dt dt = ds / v ∫ 0 t dt =∫ 0 s ds / v ∫ 0 t dt = (^) ∫ 0 2000 ds / (^) √( 12 s +¿ 0.02 s 2 )¿ Solving for t we get: t = 19.27s
Question # 12.
A particle is moving with a velocity of v 0 when s = 0 and t = 0. If it is subjected to a deceleration of a = -kv^3 , where k is a constant, determine its velocity and position as time functions.
Solution Question # 12.
Expressing the velocity of a particle as a function of time dv = adt dv = (-kv^3 ) dt dv /v^3 = (-k) dt ∫ v 0 v dt / v 3 =− k ∫ 0 t dt (-1/2) [v-2^ – v 0 -2] = -kt v (t) = sqrt [1/ {2kt + (1/v 0 2)}] Expressing the position of a particle as a function of time
Question # 12.
When a rocket reaches an altitude of 40 m it begins to travel along the parabolic path (y- 40)^2 = 160x, where the coordinates are measured in meters. If the component of velocity in the vertical direction is constant at vy = 180 m/s, determine the magnitudes of the rocket’s velocity and acceleration when it reaches an altitude of 80 m.
Solution Question # 12.
Given: h 1 = 40m, h 2 = 80m, v 0 = 180m/s, b = 160m y - h 12 = bx 2(y - h 1 ) vy = bvx bax = 2vy^2 vx2 = (2/b) (h 2 – h 1 ) v 0 vx2 = (2/160) (80 – 40) (180) = 240 m/s ax2 = (2/b) v 02 ax2 = (2/160) (180)^2 = 405 m/s^2
vy2 = v 0 = 180 m/s v 2 = sqrt (vx2^2 + vy2^2 ) v 2 = sqrt (240^2 + 180^2 ) = 201.246 m/s ay2 = 0 a 2 = sqrt (ax2^2 + ay2^2 ) a 2 = sqrt (0^2 + 405^2 ) = 405 m/s^2
Question # 12.
The motorcycle travels with constant speed v0 along the path that, for a short distance, takes the form of a sine curve. Determine the x and y components of its velocity at any instant on the curve.
Solution Question # 12.
Consider the equation for the position of the motorcycle on the sinusoidal portion of the rod: y = c sin(πx/L) Differentiate y with respect to time t. vy = dy/dt vy = (dy/dx) (dx/dt) vy = (d/dx) (c sin(πx/L)) (dx/dt) vy = (cπ/L) cos (πx/L) (dx/dt)
Sy is the displacement of the ball in y direction, Sy = h 2 - h 1 Sy = h 2 – h 1 Sy = 3m - 2m Sy = 1m uy is the initial velocity of the ball in y direction uy = VASin 30 uy = 0.5VA t is the time taken by the ball from the point of the projection ay is the acceleration due to gravity in the -ve y direction Now Equation 3 becomes: 1 = 0.5VAt – (1/2) (9.8) t^2 1 = 0.5VAt – 4.5t^2 ----- (4) Now we have R = VA cos 30t 10 = VA 0.87t t = 10 / 0.87VA Putting the value of t in Equation 4 1 = 0.5VA (10/0.87VA) – 4.5(10/0.87VA)^2 0 = -1 + 0.5VA (10/0.87VA) – 4.5(10/0.87VA)^2 Solving for VA we get: VA = 11.68 m/s Applying the equation of y motion for the basketball, Vy^2 – uy^2 = 2aySy
Vy^2 = (0.5VA)^2 - 2 x 9.8 × 1 Vy^2 = (0.5*11.68)^2 - 2 x 9.8 × 1 Vy^2 = 14. Vy = 3.8 m/s Vx = VA cos 30 Vx = 11.68 cos 30 Vx = 10.1 m/s The velocity of the ball when it passes through the basket is found by: VB = sqrt (Vx^2 + Vy^2 ) VB = sqrt (10.1^2 + 3.8^2 ) VB = 10.8 m/s
Question # 12.
A projectile is given a velocity of v 0 at an angle ϕ above the horizontal. Determine the distance d to where it strikes the sloped ground. The acceleration due to gravity is g.
Solution Question # 12.
Consider the diagram as shown in Figure 1. Figure 1 : Projectile Motion
Solution Question # 12.
xA = 0m x B = 30m vx = 30 cos ϴ yA = 0m yB = -1.2m vy = 30 sin ϴ a = -g = -9.81 m/s^2 x = x 0 + (v 0 ) (^) x t 30 = 0 + (v cos ϴ) t t = 1 / cos ϴ y = y 0 + (v 0 ) (^) y t + 0.5 act^2 -1.2 = 0 + 30 sin ϴ + 0.5 (9.81) t^2 -4.905 t^2 + 30 sin ϴ t + 1.2 = 0 -4.905 (1 / cos ϴ)^2 + 30 sin ϴ (1 / cos ϴ) + 1.2 = 0 -4.905 + 30 sin ϴ cos ϴ + 1.2 cos^2 ϴ = 0 ----- (5) Solving Equation 5 we get:
