Question

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.

Answer 1

To find the answers, we can use the formulas of motion with constant acceleration. With the given information, we can substitute the values into these formulas to find the answers. To graph x versus t and v versus t, we can plot the values of distance and velocity against time on a coordinate plane.

Step-by-step explanation:

To find answers to the given questions, we can use the formulas of motion with constant acceleration:

(a) Acceleration (a) = Change in velocity (Δv) / Change in time (Δt)

(b) Time (Δt) = Change in distance (Δd) / Velocity (v)

(c) Time (Δt) = Velocity (v) / Acceleration (a)

(d) Distance (d) = (1/2) * Acceleration (a) * Time^2 (Δt^2)

Based on the given information, we can substitute the values into these formulas to find the answers.

(a) Acceleration (a) = (50 m/s - 30 m/s) / Δt

(b) Δt = 160 m / 30 m/s

(c) Δt = 30 m/s / a

(d) Distance (d) = (1/2) * a * (Δt^2)

(e) To graph x versus t and v versus t, we can plot the values of distance and velocity against time on a coordinate plane.

Answer 2

a) 5m/s2

b) 4 sec

c) 6 sec

d) 90 m

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

Step-by-step 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