Which of the following 1D trajectories has a point t >0 where the velocity is O but the acceleration is not? O x(t) = -4t +1 O x(t) = In t O x(t) = 3e*/2 O x(t) = 4t + 9.8t2
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- An inquisitive physics student and mountain climber climbs a 51.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.06 m/s. (a) How long after release of the first stone do the two stones hit the water?Derive the equation for velocity and acceleration given that position x = v0t + 1/2 at2, where v0and a are the initial velocity and constant acceleration, respectively. Find the position.An inquisitive physics student and mountain climber climbs a 51.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.06 m/s. (b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?
- A particle moving along a straight line will be s centimetres from a fixedpoint at time t seconds, where t > 0 and s = 27t^3 + 16/t + 10a. Determine when the velocity will be zero.b. Is the particle accelerating? Explain.If acceleration is a function of displacement, integrate a ds = v dv to obtain the velocity as a function of position. TRUE OR FALSE Ineed asap thank youThe position of a particle along a straight-line path is defined by s = (t3 - 6t2 - 15t + 7) ft, where t is in seconds. Determine: (a) the total distance traveled when t = 15s (b) What are the particle’s average velocity, average speed, and the instantaneous velocity and acceleration at this time?
- A particle is moving with a velocity of 60m/s in the positive x-direction at t = 0s. Between t = 0s andt = 15s, the velocity decreases uniformly to zero. What was the acceleration during this 15s interval? Whatis the signi cance of the sign of your answer?a parachutist opens his chute at time t=0, while falling vertically at a speed of 50.0 m/s. From the combined effects of gravity and air resistance, he is subject to anupward acceleration of ay= (40/T)e-t/T where T=6.8s. Find the parachutist's speed 10s after the chute is opened and the distance fallen during that period of time.Given the velocity of an object and its initial position, how do you find the position of the object, for t ≥ 0?
- A particle moves in a straight line and had acceleration given by a(t) = -3t - 5 m/s^2. It’s initial velocity is u(0) = - 5 m/s and it’s initial displacement is s(0) = 1m. Find it’s position function s(t).Find the Velocity and the acceleration of the following Position r = A(exp(αt/2)ˆi + exp(-αt/2) ˆj), where A and α are constants.The acceleration of a particle as it moves along a straight line is given by a = (2t - 1) m/s^2 , where t is in seconds. If s = 1 m and v = 2 m>s when t = 0, determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during thistime period.