By Resnick, Halliday and Walker
Read or Download Instructor Solution Manual for Fundamentals of Physics 9thEd Resnick, Walker, and Halliday PDF
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Extra resources for Instructor Solution Manual for Fundamentals of Physics 9thEd Resnick, Walker, and Halliday
Example text
5 s . 2 m / s2 (b) We take the car to be at x = 0 when the brakes are applied (at time t = 0). 2 m/s2 ) t 2 2 in SI units. 5 s on the graph to the right. We have not shown the v-vs-t graph here; it is a descending straight line from v0 to v. 31. The constant acceleration stated in the problem permits the use of the equations in Table 2-1. 0 × 10 8 m / s) = 31 . 2 months. (b) We evaluate x = x0 + v0 t + 12 at 2 , with x0 = 0. 6 ×1013 m . ( 2 Note that in solving parts (a) and (b), we did not use the equation v 2 = v02 + 2a( x − x0 ) .
We note that the distance Δx consists of the original gap between them, D, as well as the forward distance traveled during this time by the locomotive vA t . Therefore, vt + v A Δ x D + vA t D = = = + vA . 2 t t t We now use Eq. 2-11 to eliminate time from the equation. Thus, vt + v A D = + vA v A − vt / a 2 b g which leads to F v + v − v IJ FG v − v IJ = − 1 bv − v g . 994 m/s2. A graph is shown here for the case where a collision is just avoided (x along the vertical axis is in meters and t along the horizontal axis is in seconds).
Taking the time to be zero at the instant the key is dropped, we compute the time t when y = 45 m. Since the initial velocity of the key is zero, the coordinate of the key is given by y = 12 gt 2 . 03 s . 0 m / s . 03 s 54. 8 m/s2 (taking down as the –y direction) for the duration of the motion. We are allowed to use Eq. 2-15 (with Δy replacing Δx) because this is constant acceleration motion. We use primed variables (except t) with the first stone, which has zero initial velocity, and unprimed variables with the second stone (with initial downward velocity –v0, so that v0 is being used for the initial speed).