Different Sources Of Kinematics Quiz

1. A particle with initial speed 8m/s moves in a straight line with constant acceleration. If it travels 640m in 40 seconds, its average velocity, final velocity, and acceleration, respectively, are:

4m/s, 16m/s, 0.0m/s2

2560m/s, 24m/s, 40m/s2

16m/s, 24m/s, 0.4m/s2

24m/s, 2560m/s, 0.4m/s2

Ave. vel = total_disp/time = 640m/40s = 16m/s; s= ut+½at&sup2 ==> 640 = 8*40 +½ a 40&sup2 which can be solved to give a = 0.40 m/s&sup2. Now that we know "a", we can use v = u + at to find v = 8m/s + 0.4m/s&sup2 * 40s = 24m/s

Explanation

Ave. vel = total_disp/time = 640m/40s = 16m/s; s= ut+½at&sup2 ==> 640 = 8*40 +½ a 40&sup2 which can be solved to give a = 0.40 m/s&sup2. Now that we know "a", we can use v = u + at to find v = 8m/s + 0.4m/s&sup2 * 40s = 24m/s

2. A bicyclist starts from rest and moves with a constant acceleration of 3 m/s&sup2. After 4 seconds, how far has he traveled?

24 m

48 m

12m

Not enough information

s = ut + ½ a t&sup2 = 0 * 4 + ½ * 3m/s&sup2 * (4 s)&sup2 = 24 m

Explanation

s = ut + ½ a t&sup2 = 0 * 4 + ½ * 3m/s&sup2 * (4 s)&sup2 = 24 m

3. A car accelerates uniformly and passes checkpoints A and then B. If the distance AB is 40m, determine the car's acceleration given that the time between checkpoints is 4 s and the speed of the car at point A was7. 5 m/s

30 m/s

2.5 m/s²

900.0 m/s²

1.25 m/s²

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

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

4. A particle moves such that its position along the x axis at time t is given byx= 5 + 3t -10t² while its position along the y axis is given byy= 12.5 -8t +15t² Determine its initial velocity and the time when its velocity along the x axis is zero.

(5,12.5) and t=0.267

(5,12.5) and t=0.0

(3,-8) and t= 0.267

(3,-8) and t=0.15

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

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

5. A charged particle in an electric field moves in a straight line with a constant acceleration. If its initial speed was zero and it attained a velocity of 40 km/s by travelling a distance of 2 cm, determine its constant acceleration

106 m/s²

4 X 108 m/s²

400 m/s²

200 m/s²

First convert 40 km/s to m/s: 40 X 10&sup3 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 10&sup3)^2 = 0 + 2 * a* ( 2 X 10^-2). Solve for a to get 4 X 10⁸ m/s&sup2

Explanation

First convert 40 km/s to m/s: 40 X 10&sup3 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 10&sup3)^2 = 0 + 2 * a* ( 2 X 10^-2). Solve for a to get 4 X 10⁸ m/s&sup2

6. A bullet is fired horizontally from a gun that is 50.0 m above the flat horizontal ground. If the speed of the bullet at the point of release is 200 m/s, how long does the projectile remain in the air and how far horizontally from the point of release does it land?Take g = 9.8 m/s²

3.19 s and 639 m

2.26 s and 452 m

5.10 s and 1020 m

10.20 s and 2040 m

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 at&sup2. Since there is no acceleration along the x-direction, this reduces to x = 200* 3.194 = 638.88 m

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 at&sup2. Since there is no acceleration along the x-direction, this reduces to x = 200* 3.194 = 638.88 m

7. Consider an Earth satellite that orbits 700 km above the surface of the Earth. The radius of the Earth is about 6400 km. If the period of this satellite is 100 minutes and assuming its orbit is circular, determine its speed and centripetal acceleration.

6.702 m/s and 0.007.0 m/s²

7435.1m/s and 7.8 m/s²

6702.1 m/s and 7.0 m/s²

733.0 m/s and 0.77 m/s²

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 = v&sup2/r = (7435.1)&sup2/7100000 = 7.786 m/s&sup2

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 = v&sup2/r = (7435.1)&sup2/7100000 = 7.786 m/s&sup2

8. Suppose the radius of the Earth is 6,400km. The acceleration of a person at latitude 60° due to the Earth's rotation is

0.034 m/s²

232.7 m/s²

0.0169 m/s²

465.4 m/s²

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

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

9. A bullet is fired horizontally from a gun that is 50.0 m above the flat horizontal ground. If the speed of the bullet at the point of release is 200 m/s. What is the magnitude of its velocity as it hits the ground? Take g = 9.8 m/s²

202.43 m/s

50 m/s

200 m/s

31.262 m/s

First, we find the time of flight from the vertical component starting from s= ut + ½ at&sup2 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 t&sup2 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 √[200&sup2 + (-31.262)&sup2] = 202.43m/s

Explanation

First, we find the time of flight from the vertical component starting from s= ut + ½ at&sup2 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 t&sup2 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 √[200&sup2 + (-31.262)&sup2] = 202.43m/s

10. A stone is attached to a string that is 1.0 m long. A girl whirls the stone at constant speed in a horizontal circle that is 1.5 m above the ground. Suddenly, the string breaks and hits the ground 11 m away from the girl. Determine the centripetal acceleration of the stone before the string broke. Take g=9.8 m/s&sup2

326.67 m/s²

425.92 m/s²

255.32 m/s²

311.22 m/s²

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

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