-
-
Acceleration, Initial Velocity, Final Velocity, Time
-
Acceleration Not Given
Distance, Final-Initial Velocity, Time
-
Final Velocity Not Given
Distance, Initial Velocity, Time, Acceleration
-
Time NOT Given
Final-Initial Velocity,Distance, Acceleration
-
1. A car is travelling at 35m/s. How long will it take for the car to go 5692m?
- Given: S=35m/s, D=5692m, T=?
- Formulas: S=D/T
- Work:t=5692/35, t=163s
-
2. A boy is walking at constant spped of 2.3m/s. How far will he have walked after 85.2s?
- Given: S=2.3m/s, T=85.2s, D=?
- Formulas: S=D/T
- Work:D=85.2*2.3, D=196m
-
9. a car speeds up from 13m/s to 25m/s in a time of 3.45s. What is the car's average acceleration?
- Given: Vi=13m/s, Vf=25m/s, T=3.45s, A=?
- Formulas:A=(Vf-Vi)/T
- Work: A=(25-13)/3.45, A=3.5m/s2
-
12. A train is moving along at 45mi/hr and accelerates at a rate of 3.8m/s2 for 10s. What is the train's velocity at the end of this time period?
- Given:T=10s, Vi=45mi/hr, A=3.8m/s2, Vf=?
- Formulas:A=(Vf-Vi)/T
- Work:(45mi/1hr)(1hr/3600s)(1609m/1mi)=20m/s
- 3.8=(Vf-20)/10
- Vf=58m/s
-
15. An electric race car is moving at 5.2m/s and accelerates to 8.7m/s, while covering a distance of .63m. How long was the car accelerating?
- Given: Vi=5.2m/s, Vf=8.7m/s, D=.63m, T=?
- Formulas: D=(Vi+Vf/2)T
- Work:.63=(5.2+8.7/2)t, T= .090s
-
19. a car slows down uniformly from a speed of 32m/s, at a rate of -3.4m/s2 for 5.82s. What is the car's displacement during that time?
- Given: Vi=32m/s, A=-3.4m/s2, T=5.82s, D=?
- Formulas:D=Vi(t)+ 1/2A(t^2)
- Work:D=32(5.82)+1/2(-3.4)(5.82^2)
- D= 129m
-
21. A driver is moving at 65mph slams on the brakes. If friction between the tires and the road provides a max deceleration of -8.2m/s2, what is the minimum stopping distance required?
- Given: Vi=65mph, A=-8.2m/s2, Vf=0, D=?
- Formulas: Vf^2=Vi^2+2AD
- Work:(65mi/1hr)(1hr/3600s)(1609m/1mi)
- 0=29^2+2(-8.2)(X), D= 51m
-
26. A stone is dropped from the top of a cliff. It hits the ground below 4.2s after being dropped. How high is the cliff?
- Given: T=4.2s, Vi=0, D=?, A=9.8m/s
- Formulas: D=Vi(t)+1/2AT^2
- Work:D=0+1/2(9.8)(4.2)^2
- D=86m
-
28. How long did it take for King Kong to fall straight down from the top of the Empire State Building(381m).
- Given: Vi=0, A=9.8, D=381, T=?
- Formulas: D=Vi(t)+1/2AT^2
- Work: 381=0+1/2(9.8)(x^2)
- T=8.8s
|
|