Name: ________________________
Physics I H
Date: ______________
Mr. Tiesler
Chapter 3 Homework Problems 6-10
6.) Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow,
and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the
displacement of the car during the skidding process. (Note that the direction of the velocity and the
acceleration vectors are denoted by a + and a - sign.)
𝑣0 = +30.0 𝑚/𝑠
𝑣=0
𝑎 = −8.00 𝑚/𝑠 2
𝑥 =?
𝑣 2 = 𝑣0 2 + 2𝑎𝑥
𝑥=
𝑣 2 − 𝑣0 2 0 − (+30.0 𝑚/𝑠)2
=
= 56.25 𝑚 = 56.3 𝑚
2𝑎
2(−8.00 𝑚/𝑠 2 )
7.) Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of
a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.
𝑣0 = 0
𝑎 = 6.00 𝑚/𝑠 2
𝑡 = 4.10 𝑠
𝑥 =?
1
𝑥 = 𝑣0 𝑡 + 2 𝑎𝑡 2
1
𝑥 = 0 + 2 (6.00 𝑚/𝑠 2 )(4.10 𝑠)2 = 50.43 𝑚 = 50.4 𝑚
8.) Starting from rest, a car accelerates at a constant 4.00 m/s2 for a distance of 425 m. The car is then
shifted into neutral and slows down at a rate of 2.25 m/s2. How much time elapses between when the car
starts and when it stops?
This problem has two parts. The first part is when the car is accelerating. The second part is after the
car is shifted into neutral and begins to decelerate.
To find the elapsed time for the first part, find the final velocity and then the elapsed time:
𝑣0 = 0
𝑎 = 4.00 𝑚/𝑠 2
𝑥 = 425 𝑚
To find the final velocity:
𝑣 2 = 𝑣0 2 + 2𝑎𝑥
𝑣 = √𝑣0 2 + 2𝑎𝑥 = √0 + 2(4.00 𝑚/𝑠 2 )(425 𝑚) = 58.3 𝑚/𝑠
To find the elapsed time:
𝑣 = 𝑣0 + 𝑎𝑡
𝑡1 =
𝑣−𝑣0
𝑎
=
58.3𝑚⁄𝑠−0
4.00 𝑚/𝑠2
= 14.6 𝑠
0−58.3 𝑚/𝑠
−2.25 𝑚/𝑠2
= 25.9 𝑠
To find the elapsed time for the second part:
𝑣0 = 58.3 𝑚/𝑠
𝑣=0
𝑎 = −2.25 𝑚/𝑠 2
𝑣 = 𝑣0 + 𝑎𝑡
𝑡2 =
𝑣−𝑣0
𝑎
=
To find the total elapsed time:
𝑡𝑡𝑜𝑡𝑎𝑙 = 𝑡1 + 𝑡2 = 14.6 𝑠 + 25.9 𝑠 = 40.5 𝑠
9.) A bullet in a rifle accelerates uniformly from rest at a = 7.00x104 m/s2. If the velocity of the bullet as it
leaves the muzzle is 5.00x102 m/s, how long is the rifle barrel? How long did it take for the bullet to
travel the length of the barrel?
𝑣0 = 0
𝑣 = 5.00𝑥102 𝑚/𝑠
𝑎 = 7.00𝑥104 𝑚/𝑠 2
𝑥 =?
𝑡 =?
To find the length of the barrel:
𝑣 2 = 𝑣0 2 + 2𝑎𝑥
𝑥=
𝑣 2 −𝑣0 2
2𝑎
𝑚
5.00𝑥102 −0
= 2(7.00𝑥104 𝑠𝑚/𝑠2 ) = 1.79 𝑚
To find the time for the bullet to travel the length of the barrel:
𝑣 = 𝑣0 + 𝑎𝑡
𝑡=
𝑣−𝑣0
𝑎
=
5.00𝑥102 𝑚⁄𝑠−0
7.00𝑥104 𝑚/𝑠2
= 0.00714 𝑠 = 7.14𝑥10−3 𝑠
Extra-Difficult-But-You-Can-Solve-It-If-You-Try-So-Don’t-Give-Up-Challenge-Problem!
10.) A red car is stopped at a red light. As the light turns green, it accelerates forward at 2.00 m/s 2. At the
exact same instant, a blue car passes by traveling at 62.0 km/h. When and how far down the road will the
cars again meet?
Red Car
Blue Car
𝑣0 𝑅𝑒𝑑 = 0
𝑣0 𝐵𝑙𝑢𝑒 = 62.0 𝑘𝑚⁄ℎ , 17.2 𝑚⁄𝑠
𝑎𝑅𝑒𝑑 = 2.00 𝑚⁄𝑠 2
𝑎𝐵𝑙𝑢𝑒 = 0
𝑡𝑅𝑒𝑑 =?
𝑡𝐵𝑙𝑢𝑒 =?
∆𝑥𝑅𝑒𝑑 =?
∆𝑥𝐵𝑙𝑢𝑒 =?
Since the cars meet at the same spot down the road, the displacement and the time interval is the same for
both cars.
∆𝑥𝑅𝑒𝑑 = ∆𝑥𝐵𝑙𝑢𝑒
and
𝑡𝑅𝑒𝑑 = 𝑡𝐵𝑙𝑢𝑒
1
To find displacement as a function of time, use ∆𝑥 = 𝑣0 𝑡 + 2 𝑎𝑡 2 . Since the displacements for both cars
are the same,
1
1
𝑣0 𝑅𝑒𝑑 𝑡𝑅𝑒𝑑 + 𝑎𝑅𝑒𝑑 𝑡𝑅𝑒𝑑 2 = 𝑣0 𝐵𝑙𝑢𝑒 𝑡𝐵𝑙𝑢𝑒 + 𝑎𝐵𝑙𝑢𝑒 𝑡𝐵𝑙𝑢𝑒 2
2
2
Since 𝑡𝑅𝑒𝑑 = 𝑡𝐵𝑙𝑢𝑒 , we can rewrite the equation as
1
1
𝑣0 𝑅𝑒𝑑 𝑡 + 𝑎𝑅𝑒𝑑 𝑡 2 = 𝑣0 𝐵𝑙𝑢𝑒 𝑡 + 𝑎𝐵𝑙𝑢𝑒 𝑡 2
2
2
Since 𝑣0 𝑅𝑒𝑑 = 0 and 𝑎𝐵𝑙𝑢𝑒 = 0, we can further rewrite the equation as
1
𝑎 𝑡 2 = 𝑣0 𝐵𝑙𝑢𝑒 𝑡
2 𝑅𝑒𝑑
Solving for t yields
𝑡=
2(𝑣0 𝐵𝑙𝑢𝑒 ) 2(17.2 𝑚⁄𝑠)
=
= 17.2 𝑠
𝑎𝑅𝑒𝑑
2.00 𝑚⁄𝑠 2
The value for t can now be substituted into the equation for either the red or blue car to find displacement
1
1
Red car: ∆𝑥 = 𝑣0 𝑅𝑒𝑑 𝑡 + 2 𝑎𝑅𝑒𝑑 𝑡 2 = 0 + 2 (2.00 𝑚⁄𝑠 2 ) = 295.8 𝑚 = 296 𝑚
1
Blue Car: ∆𝑥 = 𝑣0 𝐵𝑙𝑢𝑒 𝑡 + 2 𝑎𝐵𝑙𝑢𝑒 𝑡 2 = (17.2 𝑚⁄𝑠)(17.2 𝑠) + 0 = 295.8 𝑚 = 296 𝑚