5 A train started from rest and moved with constant acceleration. At one time it was traveling at 33.0 m/s, an

Question

Q.5 A train started from rest and moved with constant acceleration. At one time it was traveling at 33.0 m/s, and 160 m farther on it was traveling at 54.0 m/s. Calculate
a) the acceleration
b) the time required to travel the 160 m
c) the time required to attain the speed of 33.0 m/s
d) the distance moved from rest to the time the train had a speed of 33.0 m/s

Answer 1

Started from rest u=0
moved with constant acceleration (f). let at time t=t(33) it was moving with v-33.

[v-33]^2 = 0 + 2 f * s-33 and [v-33] = 0 + f * t-33 -----(1)
s-33 = [v-33]^2 /2 f
s-33 = (33)^2 /2f ---------------(2)

s-33 is the distance it travelled till time t-33. After 160 meter from this its speed was 54 m/s so

[v-54]^2=0 + 2 f *[ s-33+160] and [v-54] =0+ f [s-33+160]
[54]^2 = 2 f * [ s-33 + 160]
[ s-33 + 160] = [54]^2 / 2 f -----(3)

a) put the value of s-33 from (2) in (3)

[(54)^2 / 2 f] - [(33)^2 /2f] =160
[2916 - 1089] =320 f
Acceleration = f = 5.709 m/sec^2
----------------------
next part (c) first
c) [v-33] = 0 + f * t-33
t-33 = 33 / 5.709 = 5.78 sec
so it took t-33 = 5.78 secs to attain the speed of v-33
--------------------
b) the time required to travel intermediate 160 m

f = [v-54 - v-33] / [t-54 - t-33] as acceleration is constant
5.709 = [54 - 33] / [t-54 - 5.78]
[t-54 - 5.78] = 3.678
t-54 = 9.458 secs
this is the time when it had moved s-33+160 meters

time in 160 meter = t-54 - t-33 = 3.678 secs

d) s-33
s-33 = (33)^2 /2f = 1089 / 2*5.709
s-33 = 95.375 meter
all covered

Answer 2

Acceleration (a) equal to the change of velocity (v) over time (t)
a =(v2-v1)/(t2-t1)

s=0.5at^2
v2=at2
v1=at1

so s2-s1=0.5 a(t2^2-t1^2)
s2-s1=0.5 a(t2-t1)(t2+t1)

s2-s1=0.5 (t2-t1)(at2+at1)
then
s2-s1=0.5 (t2-t1)(v2+v1) now
(t2-t1)=2(s2-s1)/(v2+v1)
since
a =(v2-v1)/(t2-t1) substituting (t2-t1)
a = (v2-v1) /[2(s2-s1)/(v2+v1)]
a=(54-33)/[2 (160)/(54+33)=
a=5.71 m/sec^2

(b) t=t2-t1
t1=v1/a=33/5.71=5.8 sec
t2=v2/a=9.5sec

t=3.7 sec
(c) well this is out t1=5.8 sec.
(d) s=.5 a t1^2
s=.5at^2=0.5 x 5.7 x (5.8)^2=95.9 m

That should do it
Have fun

and then I realized that anilbakshi betat me to it. Good job anilbakshi!