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Unformatted text preview: 5 (e) This has nonzero velocity at t = 0. 2. Since the slope of the tangent to the v vs t curve, at any given time, is the acceleration, and the acceleration is increasing, we look for a curve which has increasing slope with time. The only possibilities are (a) and (e). We discard the latter because it has nonzero velocity at t = 0. 3. Since a = kt and a = dv dt , we can find the velocity as a function of time by solving the differential equation dv dt = kt . We did this in class, v ( t ) = R adt + C = R ktdt + C = 1 2 kt 2 + C . The constant is fixed by the condition that v (0) = 0, giving C = 0. So we are looking for the plot in v vs t of the function v = 1 2 kt 2 . Notice that both (a) and (e) are quadratic in t but (e) is wrong because it does not satisfy that v = 0 at t = 0. 6 6. The following position versus time graph describes the motion of Bob’s sled. In the time interval shown in the graph, how many times does Bob sled come to rest (that is, it’s instantaneous velocity vanishes)? (a) 7 4 (b) 5 (c) 1 (d) 2 (e) 4 ANS: In an x vs t graph the velocity is the slope of the tangent, at any instant t . Therefore vanishing velocity corresponds to the points on the graph where the slope van ishes: the local minima and maxima of the plot. Here is the figure, redrawn with a blob indicating each point at which the slope vanishes: 7 7. Alice is a member of the UCSD robotics club. She designs a robot which can control the acceleration which with it lifts a bucket of water, so that it won’t spill. If this acceleration starting form t = 0 is given by a = 7 t 2 , with all quantities in SI units, what is the speed at which the bucket is being lifted at t = 1 s? (a) 7 3 m/s 4 (b) 7 m/s (c) 14 m/s (d) 7 2 m/s (e) 21 m/s Since a = 7 t 2 and a = dv dt , we can find the velocity as a function of time by solving the differential equation dv dt = 7 t 2 . We did this in class, v ( t ) = R adt + C = R 7 t 2 dt + C = 7 3 kt 2 + C . The constant is fixed by the condition that v (0) = 0, giving C = 0 (that the velocity is zero at t = 0 is not stated explicitly, but here is were common sense comes in, as explained in class, the robot must start from a stationary position when it starts lifting the bucket). So we have v ( t ) = 7 3 kt 2 and v (1 s) = 7 3 m/s. Good luck! 8...
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 Winter '07
 Hicks
 Physics, Derivative, Acceleration, Velocity, Speed limit

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