motion problem? A truck starts from rest, etc.?

I

indiegirl

Member
A truck starts from rest at the top of a uniform slope and coasts down it a distance of 175 m, as a result its height above the ground is reduced by 59.1m. The plane is frictionless. What is the trucks velocity when it reaches the 175m mark, and how long does it take to reach this mark?

I really don't understand this question. Any help is appreciated, thanks. :)
 
N

NoushadBinJamal

New member
well,

good question..

you understand that the body moves under gravity.... ok?..

over a plane having friction = 0(theoretically)

distance travelled= 175m

gradient theta= tan-1(59.1/175)

theta=18.66 degree

now

friction =0

as the body moves

m*g*h* sin(theta)= 1/2 mv^2

g*h*sin(theta) = 1/2 v^2

g=accelaration due to gravity

h=59.1m
from this you can find final velocity 'v'

now for the time,

use general equation s= ut + 1/2 a t^2

v= u+ at

here u=0

so v/t = a

hence

you can calculate t=s/v

s=175 m
 
T

ThomasF

New member
first figure out slope: invsin(theta)=59.1/175
theta = 19.7 degrees

now, you break down the parallel, normal components of acceleration due to gravity, b/c friction coefficient is zero, you can ignore the normal force
sin(theta) = a(parallel)/9.81
a = 9.81*sin(19.7) = 3.31 m/s^2

now treat it as a linear problem, d=175, a=3.31, Vf=?, Vi=0

Vf^2 = Vi^2 +2*a*d
Vf^2 = 2*3.31*175
Vf=sqrt(1158.5)
Vf=34.04 m/s

for time, use a different motion equation

Vf=Vi+at
34.04 = 3.31t
t=10.3 s
 
Top