A rocket, initially at rest on the ground, accelerates straight upward with a constant acceleration of 56.0 m/s2, until the fuel is used up after 12.40 s. What is the speed of the rocket at that moment? (Neglect air resistance.)   From the time the fuel runs out, how long does it take for the rocket to reach its maximum height? (Use g  =  9.8 m/s2)______ What is the speed of the rocket when it hits the ground?________

You launch a model rocket from ground level. It moves directly upward with a constant acceleration of 83.5 m/s2 for 1.10 seconds, at which point it runs out of fuel. Assuming air resistance on the rocket is negligible, what is the maximum altitude (above the ground) achieved by the rocket?______m

The acceleration of a bus is given by a, (t) = at, where a = 1.25 m/s³ is a constant. ▾ Part A If the bus's velocity at time ₁1.15 s is 4.90 m/s, what is its velocity at time t₂ = 2.20 s? Express your answer in meters per second. = 7.10 m/s Submit ✓ Correct Correct answer is shown. Your answer 6.78 m/s was either rounded differently or used a different number of significant figures than required for this part. Part B Previous Answers If the bus's position at time ₁ = 1.158 is 6.00 m, what is its position at time (₂2.20 s? Express your answer in meters. ΓΓΙ ΑΣΦ VO z= 7.90 A Submit Previous Answers Request Answer < Return to Assignment P * Incorrect; Try Again; 8 attempts remaining You may have omitted the initial velocity of the bus. Provide Feedback ? m

You qualitatively analyzed the motion of a van earlier. Now, using the example of the ball thrown into the air, you can do a more detailed analysis of the van's motion. The table shown here includes the time and position data, with one worked example for finding acceleration. Time Position Velocity Ad Acceleration t(s) (m/s) Ar (m/s") 0.0 0.0 6.0 2.0 12 +3.0 12 4.0 36 Sample Calculation Notice that the velocity that will be plotted at t = 1.0 s is the average velocity between t = 0.0 s and t 2.0 s. The velocity that will be plotted at t 3.0 s is the average velocity between t = 2.0 s and 4.0 s. The acceleration that will be plotted at t= 2.0 s is the average acceleration between t = 1.0 s and t = 3.0s. 6.0 48 8.0 96 10.0 142 12.0 190 14.0 226 16.0 250 Adgd-d 12 m-0.0 m 2.0 s-0.0 s V = 18.0 262 Alo2 t2-lo 12 m 2.0 s %3D Analyze and Conclude 1. How well do the average and instantaneous velocities that you calculated agree with = 6.0 m each other? 36 m-12 m V= Afz4 %3D %3D 2 Separate the…