Physics

W. Dupre

Vandebilt Catholic High School

Solutions: Practice Midterm Exam

1. v_i = 20 m/s 2ad = v_f² - v_i² a = v_f -v_i / t

d = 42 m v_f² = 2ad + v_i² t = v_f -v_i / a

a = g = 9.8 m/s² v_f² = 2(9.8 m/s²)(42m) + (20m/s)² t = (34.9 m/s - 20 m/s) / 9.8 m/s²

t = ? v_f² = 1223.2 m²/s² t = 1.52 s

v_f = 34.9 m/s

2. a) The car is traveling with a constant velocity.

b) The car's velocity is decreasing (decelerating).

c) a = 0 (horizontal line on a velocity time graph indicates zero acceleration)

d) It is the slope of the line (a = -5 m/s²)

3. m = 5.2 kg 2ad = v_f² - v_i² SF = ma

v_i = 0 m/s a = v_f² /2d SF = 5.2 kg (14.4 m/s²)

v_f = 12 m/s a = (12 m/s)² / 2 (5m) SF = 74.88 N

d = 5 m a = 14.4 m/s²

SF = ?

4. This is a relative velocity problem.

Velocity of the river relative to the earth: v_re = -1.5 m/s (negative because it is headed downstream)

Velocity of the outboard relative to the earth: v_oe = ? (-13.5 m/s)

Velocity of the outboard relative to the river: v_or = ? (+/- 12 m/s)

Distance downstream: d_d = 24 300 m

Time to travel downstream: t_d = 1800 s

v_oe = d_d / t_d v_or = v_oe – v_re

v_oe = 24300 m / 1800 s v_or = -13.5 m/s – (-1.5 m/s)

v_oe = -13.5 m/s v_or = -12 m/s

Time to travel upstream: t_u = ?

v_or = v_oe – v_re v_oe = d_u / t_u

v_oe = v_or + v_re t_u = d_u / v_oe

v_oe = 12 m/s + (-1.5 m/s) t_u = 24300 m / 10.5 m/s

v_oe = 10.5 m/s t_u = 2314.3 s or 38.5 minutes

5. Solve using Newton’s third law: “For every action there is an equal but opposite reaction.”

m_c = 20 kg SF_c = SF_s

m_s = 3 kg m_ca_c = m_sa_s

a_c = 0.5 m/s² a_s = m_ca_{c /} m_s

a_s = ? a_s = (20 kg)(0.5 m/s²) / 3 kg

a_s = 3.33 m/s²

6. v₁ = 80 km/h v_total = d_total / t_total d₁ = v₁t₁

t₁ = 3 h v_total = 480 km / 8 h d₁ = (80 km/h)(3 h)

t₂ = 5 h v_total = 60 km/h d₁ = 240 km

v_total = ?

d₁ = d₂

d_total = d₁ + d₂ d_total = d₁ + d₂ = 240 km + 240 km = 480 km

7. d_x = 30 m E R² = x² + y² Tan q = y / x

d_y = 30 m S R² = (30 m)² + (30 m)² Tan q = 30 m S / 30 m E

d = ? R² = 1800 m² Tan q = 1.0

R = 42.4 m q = 45^O SoE

10. Do not worry about this problem. There will be nothing like it on the exam; however, if you are curious, relative to the ground the hobo’s displacement is 25.2 m 6.8^o NoW.

11. W = -3000 N W = mg

a = 1 m/s² m = W/g

F_T = ? ( Force of tension in the elevator cable) m = -3000 N / -9.8 m/s²

m = ? (306.1 kg) m = 306.1 kg

SF = ma SF = W + F_T

SF = (306.1 kg) ( 1 m/s²) F_T = SF - W

SF = 306.1 N F_T = 306.1 N – (-3000 N)

F_T = 3306.1 N

13. V = 100 km/h X = V Cos q Y = V Sin q

q = 240^O X = (100 km/h) Cos 240^O Y = (100 km/h) Sin 240^O

X = ? X = - 50 km/h Y = -86.6 km/h

Y = ?