# Uniform acceleration motion Problems and Solutions

Problem #1

A bus starts from rest and accelerates uniformly over a time of 4,0 seconds for a distance of 120 m. Determine the acceleration of the bus.

Given: distance d = 120 m, time t = 4,0 s, initial velocity vi = 0 m/s,
Find: acceleration bus, a?
we use formulas
d = vit + ½at2
120 = (0 m/s)(4,0 s) + ½ (a)(4,0 s)2
a = 15 m/s2

Problem #2
A car accelerates uniformly from 21,0 m/s to 35,0 m/s in 3,0 seconds. Determine the acceleration of the car and the distance traveled.

Given: time t = 4,0 s, initial velocity vi = 21,0 m/s, and finale velocity vf = 35,0 m/s,
Find: acceleration a,? and distance car, d?
we use formulas to determine the magnitude of acceleration
vf = vi + at
35 m/s = 21 m/s + a(4 s)
a = 3,5 m/s2
we use formulas to determine the magnitude of distance
(35,0 m/s)2 = (21,0 m/s)2 + 2(3,5 m/s2)(d)
1225 = 441 + 7d
784 = 7d
d = 112 m

Problem #3
A plane has a takeoff speed of 98 m/s and requires 1405 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

Given: distance d = 1405 s, initial velocity vi = 0 m/s, and finale velocity vf = 98 m/s,
Find: acceleration a,? and time plane, t?
we use formulas to determine the magnitude of acceleration
(98 m/s)2 = (0 m/s)2 + 2(a)(1405 m)
9604 = 0 + 2810a
a = 3,42 m/s2

we use formulas to determine the magnitude of time plane
vf = vi + at
98 m/s = 0 + (3,42 m/s2)t
t = 28,68 s

Problem #4
A body moving with uniform acceleration has a velocity of 12.0 cm/s when its x coordinate is 3.00 cm. If its x coordinate 2.00 s later is −5.00 cm, what is the magnitude of its acceleration?
Given: time t = 2.00 s, initial velocity vi = 12.0 cm/s, initial position xi = 3.00 cm, final position xf = -5,00 cm.
Find: acceleration a,?
we use formulas to determine the magnitude of acceleration
xf = xi + vit + ½at2
-5.00 cm = 3.00 cm + (12.0 cm/s)(2.00 s) + ½ a(2.00 s)2
-8.00 = 24 + 2a
a = -16 cm/s2

Problem #5
A jet plane lands with a velocity of 100 m/s and can accelerate at a maximum rate of −5.0 m/s2 as it comes to rest. (a) From the instant it touches the runway, what is the minimum time needed before it stops? (b) Can this plane land at a small airport where the runway is 0.80 km long?

Given: acceleration = -5.0 m/s2, initial velocity vi = 100 m/s, final velocity vf = 0, initial position xi = 0.
Find: (a) acceleration a,? and (b) final position jet, xf?
(a) we use formulas to determine the magnitude of time
vf = vi + at
0 = 100 m/s + (-5 m/s2)t
t = 20 s
The plane needs 20 s to come to a halt.
(b) The plane also travels the shortest distance in stopping if its acceleration is −5.0 m s2 . With xi = 0, we can find the plane’s final x coordinate using Eq. xf = xi + (vi + vf)/2 using t = 20 s which we got from part (a):
xf = xi + vit + ½at2
xf = 0 + ½ (100 m/s + 0)(20 s) = 1000 m = 1.0 km
The plane must have at least 1.0 km of runway in order to come to a halt safely. 0.80 km is not sufficient.

Problem #6
A truck is stopped at a stoplight. When the light turns green, it accelerates at 2.5 m/s2. At the same instant, a car passes the truck going 15 m/s. Where and when does the truck catch up with the car?
Given:
Initial position, xi, car = 0xi, truck = 0
Initial velocity, vi,car = 15 m/s, vi, truck = 0
Acceleration, acar = 0, atruck = 2.5 m/s2
The truck catches up with the car when
Position truck = position car
x0,truck + vi,truck t + ½atruckt2 = x0,car + vi,cart
0 + 0 x t + ½ x 2.5 m/s2t2 = 0 + (15 m/s)t
1.25t = 15
t = 12.0 s (time when the truck bumps into the car)
so the truck meets the car in position
xf,truck = xi,truck + vi,truckt + ½ atruckt2
= 0 + 0 x t + ½ x (2.5 m/s2)(12.0 s)2
xf, truck = 180 m

Problem 7#
A truck is traveling at 18 m/s to the north. The driver of a car, 216 m to the north and traveling south at 24 m/s, puts on the brakes and slows at 4.0 m/s2. Where do they meet?
Given:
Initial position, xi, car = 500 mxi, truck = 0
Initial velocity, vi,car = -24 m/s, vi, truck = 18 m/s
Acceleration, acar = 4.0 m/s2atruck = 0
The truck catches up with the car when
Position truck = position car
x0,car + vi,cart + ½acart2 = x0,truck + vi,truckt
216 m + (-24 m/s)t + ½ x (4.0 m/s2t2 = 0 + (18 m/s)t
500 – 42t + 2t2 = 0
t2 – 21t + 108 = 0
(t – 12)(t – 9) = 0
t = 9.0 s (time when the truck bumps into the car)
so the truck meets the car in position
xf,truck = xi,truck + vi,truckt
= x0,truck + vi,truckt
Xf,truck = 0 + (18 m/s)(9 s) = 162 m