p05_079.pdf

79. Since ( x 0 ,y 0 )=(0 , 0) and v 0 =6 . 0i, we have from Eq. 2-15

x =(6 . 0) t + 1

2 a x t 2

2 a y t 2 .

y =

These equations express uniform acceleration along each axis; the x axis points east and the y axis

presumably points north (the assumption is that the ﬁgure shown in the problem is a view from above ).

Lengths are in meters, time is in seconds, and force is in newtons.

Examination of any non-zero ( x, y ) point will su.ce, though it is certainly a good idea to check results by

examining more than one. Here we will look at the t =4 . 0 s point, at (8 . 0 , 8 . 0). The x equation becomes

8 . 0=(6 . 0)(4 . 0) + 2 a x (4 . 0) 2 . Therefore, a x =

−

2 . 0m / s 2 .The y equation becomes 8 . 0= 2 a y (4 . 0) 2 .

Thus, a y =1 . 0m / s 2 . The force, then, is

F = ma =

24i+12j

−→

(27

153 ◦ )

where the vector has been expressed in unit-vector and then magnitude-angle notation. Thus, the force

has magnitude 27 N and is directed 63 ◦ west of north (or, equivalently, 27 ◦ north of west).

−