1. A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 60.0 m past the point where the brakes were applied?
A. 22.52 m/s B. 18.4 m/s C. 15.0 m/s D. 12.1 m/s
2. A car is traveling at 24.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 30.0 m past the point where the brakes were applied?
A. 7.50 m/s B. 15.0 m/s C. 23.5 m/s D. 20.8 m/s
3. A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s2 .The car comes to a stop in a distance of 34.0 m. What was the car's speed when it had traveled 17.0 m from the point where the brakes were applied?
A. 10.9 m/s B. 14.5 m/s C. 10.7 m/s D. 21.0 m/s
4. A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the car a deceleration of 3.50 m/s2. If the car comes to a stop in a distance of 30.0 m, what was the car's original speed?
A. 10.2 m/s B. 14.5 m/s C. 2.5105 m/s D. 2.5210 m/s
5. A car traveling with velocity v is decelerated by a constant acceleration of magnitude a. It takes a time t to come to rest. If its initial velocity were doubled, the time required to stop would
A. double as well
B. Decrease by a factor of 2
C. Stay the Same
D. Quadruple.
We nned to calculate the acceleration using v^2 = v0^2 + 2*a*x..
since v = 0 then a = -v0^2/2x = -26^2/(2*120) = -2.817m/s^2
So when the car has gone 60m we find
v = sqrt(v0^2 + 2*a*x) = sqrt(26^2 + 2*(-2.817)*60) = 18.4m/s
2. Just like the one above with different numbers
3. First we find v0 using the v^2 = v0^2 + 2*a*x eqn where v = 0
we get v0 = sqrt(2*a*x) = sqrt(2*3.50*34) = 15.4m/s
So now when x = 17 we get v = sqrt(15.4^2 + 2*(-3.50)*17) = 10.9m/s
4.JUst like #3 above Solve using different numbers
5. Using v = v0 - a*t when it stops v = 0 ...so v0 = at..or t = v0/a
So if v0 is doubled then time is doubled
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