A truck covers 40.0 m in 8.60 s while smoothly slowing down to a final velocity of 2.00 m/s.
(a) Find the truck's original speed.
(b) Find its acceleration.
averagevelocity= 1/2 (Vf+Vi)
40/8.60=1/2 (2+Vi)
solve for the initial speed.
acceleration? (Vf-Vi)/time
To find the truck's original speed, we can use the equation of motion:
v = u + at
where:
v = final velocity = 2.00 m/s
u = initial velocity (unknown)
a = acceleration (unknown)
t = time taken = 8.60 s
We also need to use the equation of motion which relates displacement, initial velocity, final velocity, and acceleration:
s = ut + (1/2)at^2
where:
s = displacement = 40.0 m
u = initial velocity (unknown)
t = time taken = 8.60 s
a = acceleration (unknown)
Now, let's solve for the unknowns:
(a) Finding the truck's original speed (u):
Using the first equation of motion, rearrange it to solve for u:
u = v - at
Substituting the known values:
u = 2.00 m/s - a(8.60 s)
(b) Finding the acceleration (a):
Using the second equation of motion, rearrange it to solve for a:
s = ut + (1/2)at^2
Substituting the known values:
40.0 m = (u)(8.60 s) + (1/2)(a)(8.60 s)^2
Now, we have two equations and two unknowns. We can solve these simultaneous equations to find the values of u and a.