Just a homework problem I can't get...If you could show how you get the answers that would be useful in helping me actually understand it. Thanks a ton!!! All help is appreciated
A truck covers 40.0 m in 8.75 s while smoothly slowing down to final speed 2.55 m/s.
Find the A) Orginial Speed ?m/s and the B) Acceleration ?m/s2
A truck covers 40.0 m in 8.75 s while smoothly slowing down to final speed 2.55 m/s.
Find the A) Orginial Speed ?m/s and the B) Acceleration ?m/s2
-
we know that dist = avg velocity / time
this tells us that avg velocity = dist/time = 40m/8.75s = 4.57m/s
we also know that avg vel = (vf+v0)/2
we know vf=2.55m/s, so that v0 is
v0 = 2*avg vel -vf = 2*4.57-2.55=6.59m/s
a=(vf-v0)/t = (2.55m/s - 6.59m/s)/8.75s =- 0.46m/s/s
this tells us that avg velocity = dist/time = 40m/8.75s = 4.57m/s
we also know that avg vel = (vf+v0)/2
we know vf=2.55m/s, so that v0 is
v0 = 2*avg vel -vf = 2*4.57-2.55=6.59m/s
a=(vf-v0)/t = (2.55m/s - 6.59m/s)/8.75s =- 0.46m/s/s
-
Original speed u = distance / time = 40.0m / 8.75s = 4.57 m/s (2d.p.)
Use v = u - at (v = final velocity, u = initial velocity, a = acceleration, t = time)
Acceleration a = (v - u)/t = (2.55m/s - 4.57m/s)/ 8.75s = -0.231 m/s^2 (negative denotes the deceleration)
Use v = u - at (v = final velocity, u = initial velocity, a = acceleration, t = time)
Acceleration a = (v - u)/t = (2.55m/s - 4.57m/s)/ 8.75s = -0.231 m/s^2 (negative denotes the deceleration)
-
constant acceleration" a" ( 'slowing smoothly)
distance D = 1/2 a t^2 + Vot + Do
D given as 40 m. t given as 8.75 sec, assume Do = 40M and final D = O
or easier Do = 0 and D = 40m
be careful of the sign of a
solve by algebra or guess and calculate , guess again
distance D = 1/2 a t^2 + Vot + Do
D given as 40 m. t given as 8.75 sec, assume Do = 40M and final D = O
or easier Do = 0 and D = 40m
be careful of the sign of a
solve by algebra or guess and calculate , guess again