# Uniform acceleration motion Problems and Solutions

**Problem #1**

*Answer*:

Given: distance d = 120 m, time t = 4,0 s, initial velocity vi = 0 m/s,

Find: acceleration bus, a?

we use formulas

*d*=

*v*+ ½

_{i}t*at*

^{2}

120 = (0 m/s)(4,0 s) + ½ (

*a*)(4,0 s)

^{2}

*a***= 15 m/s**

^{2}**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.

Answer:

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

*v*

_{f}=

*v*

_{i}+

*at*

35 m/s = 21 m/s +

*a*(4 s)

*a***= 3,5 m/s**

^{2}we use formulas to determine the magnitude of distance

*v*

_{f}

^{2}=

*v*

_{i}

^{2}+ 2

*ad*

(35,0 m/s)

^{2}= (21,0 m/s)

^{2}+ 2(3,5 m/s

^{2})(

*d*)

1225 = 441 + 7

*d*

784 = 7

*d*

*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.

*Answer*:

Given: distance

*d*= 1405 s, initial velocity vi = 0 m/s, and finale velocity v

_{f}= 98 m/s,

Find: acceleration a,? and time plane,

*t*?

we use formulas to determine the magnitude of acceleration

*v*

_{f}

^{2}=

*v*

_{i}

^{2}+ 2

*ad*

(98 m/s)

^{2}= (0 m/s)

^{2}+ 2(

*a*)(1405 m)

9604 = 0 + 2810

*a*

*a*=

**3,42 m/s**

^{2}we use formulas to determine the magnitude of time plane

*v*

_{f}=

*v*

_{i}+

*at*

98 m/s = 0 + (3,42 m/s

^{2})

*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?

*Answer*:

Given: time

*t*= 2.00 s, initial velocity v

_{i}= 12.0 cm/s, initial position

*x*

_{i}= 3.00 cm, final position x

_{f}= -5,00 cm.

Find: acceleration a,?

we use formulas to determine the magnitude of acceleration

*x*

_{f}= x

_{i}+

*v*+ ½

_{i}t*at*

^{2}

-5.00 cm = 3.00 cm + (12.0 cm/s)(2.00 s) + ½

*a*(2.00 s)

^{2}

-8.00 = 24 + 2

*a*

*a***= -16 cm/s**

^{2}

^{}**Problem #5**

A jet plane lands with a velocity of 100 m/s and can accelerate at a maximum rate of −5.0 m/s

^{2}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?

*Answer*:

Given: acceleration

*a*= -5.0 m/s

^{2}, initial velocity v

_{i}= 100 m/s, final velocity

*v*

_{f}= 0, initial position x

_{i}= 0.

Find: (a) acceleration a,? and (b) final position jet,

*x*

_{f}?

(a) we use formulas to determine the magnitude of time

*v*

_{f}=

*v*

_{i}+

*at*

0 = 100 m/s + (-5 m/s

^{2})

*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 x

_{i}= 0, we can find the plane’s final x coordinate using Eq.

*x*

_{f}= x

_{i}+ (

*v*+

_{i}*v*

_{f})/2 using t = 20 s which we got from part (a):

x

_{f}= x

_{i}+ v

_{i}t + ½at

^{2}

x

_{f}= 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/s

^{2}. At the same instant, a car passes the truck going 15 m/s. Where and when does the truck catch up with the car?

*Answer*:

Given:

Initial position, x

_{i, car}= 0

_{, }x

_{i, truck}= 0

Initial velocity,

*v*

_{i,car}= 15 m/s,

*v*

_{i, truck}= 0

Acceleration,

*a*

_{car}= 0,

*a*

_{truck}= 2.5 m/s

^{2}

The truck catches up with the car when

Position truck = position car

x

_{0,truck}+ v

_{i,truck }t + ½a

_{truck}t

^{2}= x

_{0,car}+

*v*

_{i,car}

*t*

0 + 0 x

*t*+ ½ x 2.5 m/s

^{2}

*t*

^{2}= 0 + (15 m/s)

*t*

1.25

*t*= 15

*t***= 12.0 s**(time when the truck bumps into the car)

so the truck meets the car in position

*x*

_{f,truck}=

*x*

_{i,truck}+

*v*

_{i,truck}

*+ ½*t

*a*

_{truck}

*t*

^{2}

= 0 + 0 x

*t*+ ½ x (2.5 m/s

^{2})(12.0 s)

^{2}

**x**

_{f, 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/s

^{2}. Where do they meet?

*Answer*:

Given:

Initial position, x

_{i, car}= 500 m

_{, }x

_{i, truck}= 0

Initial velocity,

*v*

_{i,car}= -24 m/s,

*v*

_{i, truck}= 18 m/s

Acceleration,

*a*

_{car}= 4.0 m/s

^{2},

*a*

_{truck}= 0

The truck catches up with the car when

Position truck = position car

x

_{0,car}+ v

_{i,car}t + ½a

_{car}t

^{2}= x

_{0,truck}+

*v*

_{i,truck}

*t*

216 m + (-24 m/s)

*t*+ ½ x (4.0 m/s

^{2})

*t*

^{2}= 0 + (18 m/s)

*t*

500 – 42

*t*+ 2

*t*

^{2}= 0

*t*

^{2}– 21

*t*+ 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

*x*

_{f,truck}=

*x*

_{i,truck}+

*v*

_{i,truck}

t

= x

_{0,truck}+

*v*

_{i,truck}

*t*

*X*

_{f,truck}= 0 + (18 m/s)(9 s) =

**162 m**

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