Motion with constant acceleration problems and solutions

 Problem #1

A particle moves along the x-axis with an initial velocity of 5 m/s and constant acceleration. After 2 seconds, its velocity is 12 m/s. How far did it travel during this interval

Answer:
Given: initial velocity v_i = 5 m/s, final velocity v_f = 12 m/s, Δt = 2 s
Unknown: Δx = ?
Since we know v_i, v_f, and Δt, and need to know Δx, but a is not specifically given,
Δx = ½ (v_i + v_f)t
= 0.5(5 + 12) x 2 = 0.5 x 17 x 2 = 17 m

Problem #2
A car is initially moving at 10 m/s and accelerates at a constant rate of 2 m/s² for 4 seconds, in a straight line. How far did the car travel during this time?

Answer:
Given: initial velocity v_i = 10 m/s, acceleration a = 2 m/s², inteval time Δt = 4 s
Unknown: Δx = ?
The value that is not given or asked for is v, so we must use the equation that does not contain v.
x = x_i + ¹/₂at² + v_it
   = 0  + 0.5 x 2 x 4² + 10 x 4
x = 16 + 40 = 56 m
Note that since we are not given an initial position and are asked for the displacement, not the final position, we simply assume that the initial position is 0, which makes the displacement equal to the final position.

Problem #3
A rock is dropped from a cliff that is 80 m above the ground. If the rock hits the ground with a velocity of 40 m/s, what acceleration did it undergo?

Answer:
Since the rock is dropped, we know that it began at rest, so the initial velocity is 0.
Given: initial velocity v_i = 0, displacement Δx = 80 m, final velocity v_f = 40 m/s
Unknown: a = ?

The value that is neither given nor asked for is Δt, so we must use the equation which does not contain Δt. v² = v_i² + 2a(x - x_i)     = 2a(x - x_i) = 2aΔx
a = (v²)/(2Δx)    = (40²)/(2 x 80) a = 1600/160 = 10 m/s², toward the ground.

Problem #4An object starts from rest and uniformly accelerates at a rate of 2 𝑚/𝑠² for 5.0 seconds. (a) What is the object’s displacement during this 5.0 second time period? (b) What is the object’s final velocity? (c) How many seconds does it take the object to have a displacement of 15 meters? (d) Suppose the object had an initial velocity of −3 𝑚/𝑠, how would the object’s displacement compare the displacement found in part (a)? 

Answer:Given : acceleration a = 2 m/s², time t = 5.0 s, initial velocity v_i = 0, then
(a) the object’s displacement during this 5.0 second time period is
∆x = v_it + ½ at²  
 = 0 + ½ (2.0)(5.0)²∆x
= 25.0 m

(b) the object’s final velocity is
v_f = v_i + at = 0 + 2.0 x 5.0 = 10.0 m/s

(c) seconds does it take the object to have a displacement of 15 meters is

∆x = v_it + ½ at²
15 = 0 + ½ (2.0)t²
t = 3.87 s

(d) The displacement would be a smaller value than that found in part a because the object would have an initial negative displacement until it reached a positive velocity. In part a the object’s displacement was always in the positive direction.

Problem #5
A train is traveling down a straight track at 20 𝑚/𝑠 when the engineer applies the brakes, resulting in an acceleration of −1.0 𝑚/𝑠² as long as the train is in motion. How far does the train move during a 40 s time interval starting at the instant the brakes are applied?

Answer
Given: initial velocity v_i = 20 m/s, acceleration a = –1.0 m/s², time t = 40 s.  Using the uniformly accelerated motion equation ∆x = v_it + ½ at² for the full 40 s interval yields
∆x = (20)(40) + ½ (–1)(40)² = 0,

which is obviously wrong.The source the error is found by computing the time required for the train to come to rest.

This time is

t = (v_f – v_i)/t = (0 – 20)/(–1) = 20 sthus, the train is slowing down for the first 20 s and is at rest for the last 20 s the 40 s interval.

The acceleration is not constant during the full 40 s. It is, however, constant during the first 20 s as the train slows to rest. application of ∆x = v_it + ½ at² to this interval gives stopping distance as
∆x = 20 x 20 + ½ (–1)(20)² = 200 m

Problem #6
The two cars then moved together close to each other with constant speeds of v_A = 40 m / s and v_B = 60 m/s. Determine:
(a) Distance of car A from the place of departure when passing by car B
(b) The time needed for the two cars to pass each other
(c) Distance of car B from the place of departure when passing by car Aficer’s car has a greater top speed than the robbers’ car.

Answer:
(a) The travel time of car A is the same as the travel time of car B, because it departs simultaneously. Distance from A when meeting for example x, so the distance from B
 (1200 – x)t_A = t_B
s_A/v_A = s_B/v_B
(x)/40 = (1200 – x)/60
6x = 4(1200 – x)
6x = 4800 – 4x
10x = 4800x = 480 meters

(b) The time needed for the two cars to pass each other
x = v_A x t
480 = 40t
t = 12 seconds

(c) Distance of car B from the place of departure when passing by car A
s_B = v_B x
t = (60)(12) = 720 m

Problem #7
A car first moves at a speed of 15 m/s along a straight track. Later driver reduce the speed of the car with a constant deceleration of 2.5 m/s². Calculate the distance traveled by the car during a slowdown until it finally stops!

Answer:Given: intial velocity v_i = 15 m/s, final velocity v_f = 0, and acceleration a = –2.5 m/s². then, the distance traveled by the car is
v_f² = v_i² + 2a∆x0 = (15)² + 2(–2.5)∆x
∆x = 45 m 

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