Starting from rest, a car accelerates at 2.5m/s2 up a hill that
is inclined 5.1 ? above the horizontal.
a) How far horizontally has the car traveled in 14s ?
b) How far vertically has the car traveled in 14s ?
ax = 2.5 cos5.1 = 2.49 m/s^2
ay = 2.5 sin5.1 = 0.22 m/s^2
sx = 1/2 * axt^2
= 1/2 * 2.49*14^2 = 244.02 m
sy = 1/2 * 0.22*14^2 = 21.56 m(answer a)
s = sqrt(244.02^2 + 21.56^2) = 244.97 m (answer a)
a) vertically
s=1/2at^2 sin thetha
s=sin(5.1)*.5*2.5*14^2
=226.8m
b) Hrizontally
s=1/2at^2 cos thetha
Cos[5.1]*.5*2.5*14^2
=92.6 horizontal
here,
a = 2.5 m/s^2
t = 14 s
using second equation of motion ,
d= ut + 0.5 * at^2
d = 0*14 + 0.5 * 2.5 *14^2
d = 254 m
a)
Horizontal distance = 254 * cos(5.1)
Horizontal distance =244 m
car has travelled 244 m horizontally .
b )
vertical distance = 254*sin(5.1)
vertical distance = 22.6 m
it will travel 22.6 m vertically
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