Physics 2C homework
Section 2C
5. A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3.2 km in 3.5 min. How fast is the car moving after this time?
3.5 min = 210 s.
Average speed = 3 200/210 = 15.238 m/s = ½ (v + vo). vo = 6.4 m/s.
Solving for v, v = 2 × 15.238 – 6.4 = 24.076 m/s.
2D
1. A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92m/s2 for 3.6 s. Find the final speed and the displacement of the car during this time.
Vf= Vi + at= 9.92m/s
D x= Vot+ at2/ 2=.00253 km =~ .003 km
2. A car starts from rest and travels for 5.0s with a uniform acceleration of -1.5m/s2. What is the final velocity of the car? How far does the car travel in this time interval? vf = vi + a(Dt) or vf = (0 m/s) + (-1.5 m/s2)(5.0 s) = -7.5 m/s Dx = vi(Dt) + 1/2a(Dt)2= Dx = (0)(5.0 s) + 1/2(-1.5 m/s2)(5.0 s)2 = -18.75 m or -19 m (sig figs).
2E
1. Find the velocity after the stroller has traveled 6.32 m. (A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500m/s2. What is the velocity of the stroller after it has traveled 4.75m?)
Vf=2.18 m/s
3. A car accelerates uniformly in a straight line from rest at the rate of 2.3m/s2.
a. What is the speed of the car after it has traveled 55m? 15.9m/s
b. How long does it take the car to travel 55m? 6.92s
dist = 1/2 a t^2 or t =sqrt[2d/a] = sqrt[2*55m/2.3m/s/s] = 6.92s
velocity = initial velocity + a t
since initial velocity =0, velocity = 2.3 m/s/s x 6.92 s = 15.9m/s
5. An aircraft has a liftoff speed of 120 km/h. What minimum uniform acceleration does this require if the aircraft is to be airborne after a takeoff run of 240m? vf2 = vi2 + 2a(Dx), where: Dx = 240 m vi = 0 m/s Dt = ??? vf = 120 km/h or 33.3 m/s (OH, km/h to m/s is to multiply by 1000/3600) a = ??? m/s2
(33.3 m/s)2 = (0 m/s)2 + 2(??? m/s2)(240 m) so a = 2.31 m/s2
5. A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3.2 km in 3.5 min. How fast is the car moving after this time?
3.5 min = 210 s.
Average speed = 3 200/210 = 15.238 m/s = ½ (v + vo). vo = 6.4 m/s.
Solving for v, v = 2 × 15.238 – 6.4 = 24.076 m/s.
2D
1. A car with an initial speed of 23.7 km/h accelerates at a uniform rate of 0.92m/s2 for 3.6 s. Find the final speed and the displacement of the car during this time.
Vf= Vi + at= 9.92m/s
D x= Vot+ at2/ 2=.00253 km =~ .003 km
2. A car starts from rest and travels for 5.0s with a uniform acceleration of -1.5m/s2. What is the final velocity of the car? How far does the car travel in this time interval? vf = vi + a(Dt) or vf = (0 m/s) + (-1.5 m/s2)(5.0 s) = -7.5 m/s Dx = vi(Dt) + 1/2a(Dt)2= Dx = (0)(5.0 s) + 1/2(-1.5 m/s2)(5.0 s)2 = -18.75 m or -19 m (sig figs).
2E
1. Find the velocity after the stroller has traveled 6.32 m. (A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500m/s2. What is the velocity of the stroller after it has traveled 4.75m?)
Vf=2.18 m/s
3. A car accelerates uniformly in a straight line from rest at the rate of 2.3m/s2.
a. What is the speed of the car after it has traveled 55m? 15.9m/s
b. How long does it take the car to travel 55m? 6.92s
dist = 1/2 a t^2 or t =sqrt[2d/a] = sqrt[2*55m/2.3m/s/s] = 6.92s
velocity = initial velocity + a t
since initial velocity =0, velocity = 2.3 m/s/s x 6.92 s = 15.9m/s
5. An aircraft has a liftoff speed of 120 km/h. What minimum uniform acceleration does this require if the aircraft is to be airborne after a takeoff run of 240m? vf2 = vi2 + 2a(Dx), where: Dx = 240 m vi = 0 m/s Dt = ??? vf = 120 km/h or 33.3 m/s (OH, km/h to m/s is to multiply by 1000/3600) a = ??? m/s2
(33.3 m/s)2 = (0 m/s)2 + 2(??? m/s2)(240 m) so a = 2.31 m/s2