A train started from rest and moved with constant acceleration. At one time it was traveling 30 m/s, and 160 m farther on it was traveling 50 m/s.
Calculate
(a) the acceleration,
(b) the time required to travel the 160 m mentioned,
(c) the time required to attain the speed of 30 m/s,
(d) the distance moved from rest to the time the train had a speed of 30 m/s.
(e) Graph x versus t and v versus t for the train, from rest.

a)5m/s2

b)4 sec

c)6 sec

d)90 m

e)     Answer in the file attached as it is a graph

Explanation:

This question can be solved using equations of motion. The two equations are:

2(a)(s) = v² – u²

v = u + (a)(t)  

a = acceleration

s = distance

v = final velocity

u = initial velocity  

t = time

a)Given final velocity = 50m/s and initial velocity = 30 m/s for 160m journey

Using 2(a)(s) = v² - u²

2(a)(160) = 50² – 30²

320 (a) = 2500 - 900

a = 1600/320

a = 5m/s²

b)The acceleration remains constant throughout so we can use it in this part as well.

Using v = u + (a)(t)

50 = 30 + (5)(t)

t = 4 sec

c)The acceleration remains constant throughout so we can use it in this part as well. Now the initial velocity will be 0m/s and final will be 30m/s

Using v = u + (a)(t)

30 = 0 + 5(t)

t = 6 sec

d)The acceleration remains constant throughout so we can use it in this part as well. Now the initial velocity will be 0m/s and final will be 30m/s

Using 2(a)(s) = v² - u²

2(5) (s) = 30² – 0²

10(s) = 900

s = 90 m

e) Graphs are attached as image

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0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr