Different Sources Of Kinematics Quiz

Explanation
Ave. vel = total_disp/time = 640m/40s = 16m/s; s= ut+½at² ==> 640 = 8*40 +½ a 40² which can be solved to give a = 0.40 m/s². Now that we know "a", we can use v = u + at to find v = 8m/s + 0.4m/s² * 40s = 24m/s

Explanation
s = ut + ½ a t² = 0 * 4 + ½ * 3m/s² * (4 s)² = 24 m

Explanation
s = ut + ½ a t². With s= 40m, t = 4s and u = 7.5m/s, we get 40 = 7.5*4 + ½* a* 4*4 ==> a =1.2m/s²

Explanation
First convert 40 km/s to m/s: 40 X 10³ m/s. Also convert 2 cm to meters: 2 X 10-2 m. Then use v^2 = u^2 + 2*a*s with u =0 to get (4 X 10³)^2 = 0 + 2 * a* ( 2 X 10-2). Solve for a to get 4 X 108 m/s²

Explanation
Differentiate with respect to time to get V_x = 3 -20t and V_y =-8+30t. Evaluate these at t=0 to get (3,-8). Secondly, set V_x to zero to get t=3/20 = 0.15

Explanation
Fired horizontally ⇒ initial vertical velocity is zero. Use s = ut + ½ a t² with u=0, s= -50m and a=-9.8 m/s² to get t = √ [2*50m /9.8 m/s²] = 3.194 s
In order to determine the horizontal distance moved in this time, we use s = ut + ½ at². Since there is no acceleration along the x-direction, this reduces to x = 200* 3.194 = 638.88 m

Explanation
First, we find the time of flight from the vertical component starting from s= ut + ½ at² with s=-50m, and u=0 (initial velocity along y direction is zero since it was fired horizontally). This gives -50 = ½ X -9.8 X t² from which we get t=3.19s. Next, use v = u + at to determine the velocity along the y-direction: V_y = 0 + (-9.8)*3.19 = -31.262 m/s. Thus, the velocity is (200, -31.262) and the required magnitude is √[200² + (-31.262)²] = 202.43m/s

Explanation
The radius of the circle at that latitude is r=R* cos(60°) = 3200 km. The period of rotation of the Earth is 24 hours. Thus, the angular speed is &omega = 2&pi/(24*60*60) = 7.27 X 10-5 rad/s. Thus, the centripetal acceleration is r&omega² = 3200X10³ m X (7.27 X 10-5 rad/s)² = 0.0169 m/s²

Explanation
Time taken to hit the ground is obtained from s = ut + ½ at² with u=0 (i.e., no initial vertical speed), s=-1.5m and g=-9.8m/s². Thus, -1.5 = 0 + ½ X (-9.8)Xt² which gives us that t = 0.553 s. Next, since the stone hit 11 m away from the girl and the string is 1.0m long, then the string hit at (11-1)m = 10m away, horizontally, from its point of release. We can calculate its horizontal speed at point of release from s = ut + ½ at² but with a=0 (no horizontal acceleration) with s=10m and t=0.553 s which means 10m = u * 0.553s i.e., u = 10/0.533=18.07 m/s. Finally, we find the centripetal acceleration A from A= v²/r with v=18.07m/s that we got before and r=1.0m (the radius of the horizontal circle) to get 326.67 m/s²

Explanation
The radius from the center of the Earth is (6400 + 700 ) km =7100 km&omega = 2&pi/T (where T=period) and v = R &omega = 7100000 m * 2&pi/(100*60) = 7435.1 m/s. For the centripetal acceleration, we can use a = v²/r = (7435.1)²/7100000 = 7.786 m/s²