Bethany F
In a car accident involving a sports car, skid marks as long as 290 m were left by the car as it decelerated to a complete stop. The police report cited the speed of the car before braking as being "in excess of 100 mph" (161 km/h). Suppose that it took 10 seconds for the car to stop. Estimate the speed of the car before the brakes were applied.
Givens: distance = 290 meters, Vf = 0, Vi = 44.7 m/s & time = 10 seconds
Answer
Given,
x - x0 = 290 m
vf = 0 m/s
t = 10 s
Solve:
Using the following kinematic equation, [1]
x -x0 = ½ (v0 + vf) t
Rearrange the terms in the kinematics equation, solve for v0.
v0 = 2 (x - x0) / t - v
Plug in the values and calculate,
= 2 (290 m/s) / 10 s - 0
= 58 m/s
Using dimensional analysis,
v0 = 58 m/s x (3600 s / 1 hr) x (1 mi / 1.6 10³ m)
Ans. v0 = 130.5 mph
Given,
x - x0 = 290 m
vf = 0 m/s
t = 10 s
Solve:
Using the following kinematic equation, [1]
x -x0 = ½ (v0 + vf) t
Rearrange the terms in the kinematics equation, solve for v0.
v0 = 2 (x - x0) / t - v
Plug in the values and calculate,
= 2 (290 m/s) / 10 s - 0
= 58 m/s
Using dimensional analysis,
v0 = 58 m/s x (3600 s / 1 hr) x (1 mi / 1.6 10³ m)
Ans. v0 = 130.5 mph
Physics help, please?Thank you so much!!!!!!!!!!!?
equiangula
1.A 500-kilogram sports car accelerates uniformly from rest, reaching a speed of 30 meters per seconds. During the 6 seconds, the car has traveled a distance of _____m.
2. A vehicle moves in a straight line with an acceleration of 8 km/h^2. BY how much does the speed change each second?
Answer
1. a = (v-v0)/t = (30-0)/6 = 5 m/s^2
x = x0+v0(t)+0.5at^2
x = 0+0+0.5(5)(6^2) = 90 m
2. acceleration is change in velocity per time...so speed changes by
8 km/h = 8/3600 = 0.002 km/s = 2 m/s
1. a = (v-v0)/t = (30-0)/6 = 5 m/s^2
x = x0+v0(t)+0.5at^2
x = 0+0+0.5(5)(6^2) = 90 m
2. acceleration is change in velocity per time...so speed changes by
8 km/h = 8/3600 = 0.002 km/s = 2 m/s
Powered by Yahoo! Answers
No comments:
Post a Comment