Question

Student Name 1. A particle confined to motion along the x axis moves with constant acceleration fromx = 2.0 m to x 8.0 m during a 2.5-s time interval. The velocity of the particle at x - 8.0 m is 2.8 m/s. What is the acceleration during this time interval? 2. The polar coordinates of a point are r=5.50 m and Angle 240°. What are the Cartesian coordinates of this point? 3. On occasion, the notation A= [A, O] will be a shorthand notation for A=Acosi+A sin If A= [15, 80°] and B = 12î-16j, what is the magnitude of A-B? 4. A particle moves in the xy plane with a constant acceleration given by 4.0j m/s. At t= 0, its position and velocity are 10im and (-2.0i+8.0 m/s, respectively. What is the distance from the origin to the particle at t 2.0 s? 5. The vector position of a particle varies in time according to the expression -3.00i-6.00t 2j, where r is in meters and t is in seconds. (a) Find an expression for the velocity of the particle as a function of time. (b) Determine the acceleration of the particle as a function of time. (c) Calculate the particle's position and velocity at t-1.00 s. 6. 4. The vector A has components x = -5 and y +7 along the x and y axes respectively. Along a set of axes rotated 90 degrees counterclockwise relative to the original axes, the vector's components are. 7. A speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100m ahead. The pilot slows the boat witha constant acceleration of 23.50 m/s2 by reducing the throttle. (a) How long does it take the boat to reach the buoy? (b) What is the velocity of the boat when it reaches the buoy? 8. (a) Calculate the height of a cliff if it takes 2.35s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s (b) How long would it take to reach the ground if it is thrown straight down with the same speed? Hint, use quadratic approach. 9. A baseball is hit so that it travels straight upward after being struck by the bat A fan observes that it takes 3.00 s for the ball to reach its maximum height Find (a) the ball's initial velocity and (b) the height it reaches. 10.Find the cross product A XB A 2i -3j+ K and B = -5i - 2j + 2k

Answer 1

Part(1)

*u = initial velocity.

*d = distance traveled = 8 - 2 = 6 m.

*t = time interval = 2.5 s.

*v = final velocity = 2.8 m/s.

*a = acceleration.

Now using the formula of kinematics,

$d = ut + \frac{1}{2}at^2 \, \, \, eqn1\\ v = u+at\\ vt = ut + at^2\,\,\, eqn2$

Subtract eqn1 from eqn2,

$vt-d = \frac{1}{2}at^2\\ a = \frac{2(vt-d)}{t^2}\\ a= \frac{2(2.8\times 2.5 -6)}{(2.5)^2}\\ a = \frac{2}{6.25}\\ a = 0.32\ m/s^2$

Part(2)

*$\theta = 240^o$.

x - cordinate = $r \cos\theta = 5.5\times \cos240^o=5.5\times \cos(180^o+60^o)=-5.5\times \cos60^o =-2.75\ m.$

y-coordinate = $r \sin\theta = 5.5\times \sin240^o=5.5\times \sin(180^o+60^o)=-5.5\times \sin60^o =-4.73\ m.$

Part(3)

$\vec{A} = 15cos80^o\hat{i}+15sin80^o\hat{j}\\ \vec{A} = 15\times 0.17\hat{i}+15\times 0.98\hat{j} \\ \vec{A} = 2.55\hat{i}+14.7\hat{j}\\ \vec{B} = 12\hat{i}-16\hat{j}\\ \vec{A}-\vec{B} = 2.55\hat{i}+14.7\hat{j} - (12\hat{i}-16\hat{j})\\ \vec{A}-\vec{B} = (-9.45)\hat{i}+30.7\hat{j}\\ \left | \vec{A}-\vec{B} \right | = \sqrt{(-9.45)^2+(30.7)^2} = 32.12$

Part(4)

*$x_0 = initial\ position\ at\ t=0, = 10\hat{i}\ m.$

*u = initial velocity = $(-2)\hat{i}+8\hat{j}\ m/s.$

*x = final position after time t = 2 s.

Now writing the equation for motion,

$x = x_0+ut+\frac{1}{2}at^2\\ = (-2)\hat{i}+8\hat{j}+ 10\hat{i}\times 2+\frac{1}{2}\times (-4\hat{j})\times 2^2\\ = 18\hat{i}\ m.$

Distance from origin after t = 2 s, = 18 m.