Kinematic Solutions:
1. Solution: The distance traveled is 393.75 meters. Note that v_0= 0 m/s, a=3.5 m/s^2, delta t= 15 s, delta x=?, v_x=?. Using the equation delta x= v_0,x(delta t)+(½)(a)(delta t)= 0(15)+(½)(3.5)(15)^2=393.75 meters.
2. Solution: First find the variables. vo= 20 mph, v=70 mph, △x=?, △t= 3 sec, a=?. To find the acceleration, use the equation v=vo+a(△ t). Plugging in gives 70=20+3a, so a=16.67 m/s^2. When I’m uniform acceleration, △x= (½)(vo+v)(△ t)= (½)(20+70)(3)=135 mi.
3. Solution: The height requested is 89.432 meters. Note that v_(0,y)= 0 m/s, a_y= -g, delta t= 4.27 s, delta y=?, v_y=?. Using the equation delta y=(v_0,y)(delta t)+(½)(a_y)(delta t)^2, then delta y=v_(0,y)(delta t)+(½)(a_y)(delta t)^2=(0)(4.27)+(½)(-g)(4.27)^2=-89.432 m. Since delta y is the final position(in the y direction) minus the initial position(in the y direction) and the height is the initial position minus the final position, the height is the additive inverse of delta y, which is therefore, 89.432 meters.
4. Solution: A cars speedometer measures speed. Velocity is a vector quantity that has both magnitude and direction. A cars speedometer only measures the magnitude of velocity, which is speed, and does not contain information about the cars direction.
5. Solution: 32.5 meters. The displacement can be solved for by finding the area under the v vs t curve, giving 15+10+7.5=32.5 m. The acceleration is 0.3 m/s^2, which can be found by taking the slope of the v vs t curve.
6. Solution: The height requested is 78.48 meters. The variables in the equation are v_(0,y)=?, v_y=0 m/s, a_y=-g, delta t= 4s, delta y=?. Using the kinematics equation
v_y=(v_(0,y))+(a_y)(delta t), 0=v_(0,y)+(a_y)(delta t), so 0=(v_(0,y))+(a_y)(delta t), so 0=v_0,y+(-g)(4), so v_(0,y)= 39.24 m/s. Thus, delta y=(v_(0,y))(delta t)+(½)(a_y)(delta t)^2, and plugging in the known quantities delta y=(39.24)(4)+(½)(-g)(4)^2= 78.48 m.
7. Solution: No, this is not necessarily true. The speed is the magnitude of the velocity and the acceleration is the change in velocity over time. It does not make sense for the change in velocity over time to be necessarily correlated with velocity. For example, an object can have a speed of 5 m/s and an acceleration of 1 m/s^2, while another object can have a speed of 8 m/s and an acceleration of 0.5 m/s^2.
8. Solution: Using relative motion, v_(0,x)=10 m/s, v_x=0 m/s, a=?, delta x= 20 m, delta t=?. Therefore, v^2=v_0^2+2a(delta x), 0^2=(10^2)+2a(20), so a= -4 m/s^2.
9. Solution: voy= vsin(θ), v(y)=0 m/s, ay=-g m/s^2, △y=?, △t=?. To solve for △t, note that vy=voy+ay(△ t). Plugging in gives 0=vsin(θ)-g(△ t). Thus, △t= vsin(θ)/g. Use the equation △y= (vy)(△ t)+(½)(ay)(△ t)^2. Plugging in gives h=(vsin θ)^2/g+(½)(-g)(vsin θ)/g= (vsin θ)^2/g-(sin θ)/2, which is in terms of the components listed.
10. Solution: The velocity at the top of projectile motion is actually the smallest at any other given point because that is when the object is in rest.
11. Solution: The horizontal velocity of the toy airplane is not affected by the change in time because acceleration in the y-direction is 0 m/s.
12. Solution: B, A, C. The horizontal velocities will be the same for all 3 scenarios so the only variable needed for consideration is the speed. Since speed and vertical displacement are proportional, a higher maximum height would mean a higher speed and, therefore, a longer flight time.
13. Solution: We know voy is equal to zero so, using the equation △y = voyt - ½ (gt^2), we get
△y = - ½ (gt^2). By plugging in -10 for △y and -9.81 for g, we find out that t ≈ 1.43 s
14. Solution: 13.216 m. Note that the velocity in the x-direction is 25cos(45) at all times because velocity in the x-direction is constant(and acceleration in the x-direction is 0 m/s^2). Since delta x is 45 m, delta t= 45/(25cos(45))= 2.546 s. Then, using the equation
delta y=(v_(0,y))(delta t)+(½)(a_y)(delta t)^2 and the facts that v_(0,y)= 25 sin(45) and a_y=-g. After plugging in this information and delta t= 2.546s, delta y= 13.216 m as claimed.
15. Solution: The initial speed given to the football was 11.523 m/s. Note that it takes 30/343=0.087 seconds for the sound of the football hitting the ground to reach Andy, so the football spends 3.9125 seconds in the air. Then, finding the variables in the y-direction yields that a_y=-g, delta y= -30 m and delta t= 3.9125 s. Using the equation
Delta y=(v_(0,y))(delta t)+(½)(a_y)(delta t)^2, it can be found that
(-30)=(v_(0,y))(3.9125)+(½)(-g)(3.9125)^2, and solving for v_(0,y) gives the answer.