Asked by Nick
A car traveling 56 km/h is 21.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.17 s later.
a)What was the car's constant deceleration before impact?
b)How fast was the car traveling at impact?
a)What was the car's constant deceleration before impact?
b)How fast was the car traveling at impact?
Answers
Answered by
Ryan
First you must change 56 km/h to m/s. In m/s the Initial Velocity is 15.56 m/s. Next use the equation:
S = ViT + (1/2)at^2 and solve for a
21 = (15.56)(2.17) + (1/2)(a)(2.17)^2
a = -5.42 m/s^2
Now that you have the acceleration you can solve for the final velocity using either:
Vf = Vi + at ---or--- Vf^2 = Vi^2 + 2as
Vf = 15.56 + (-5.42)(2.17) = 3.8 m/s
or
Vf^2 = 15.56^2 + 2(-5.42)(21) = 3.8 m/s
As you can see each equation gave the same answer. Overall your answers would be:
a.) -5.42 m/s^2
b.) 3.8 m/s
S = ViT + (1/2)at^2 and solve for a
21 = (15.56)(2.17) + (1/2)(a)(2.17)^2
a = -5.42 m/s^2
Now that you have the acceleration you can solve for the final velocity using either:
Vf = Vi + at ---or--- Vf^2 = Vi^2 + 2as
Vf = 15.56 + (-5.42)(2.17) = 3.8 m/s
or
Vf^2 = 15.56^2 + 2(-5.42)(21) = 3.8 m/s
As you can see each equation gave the same answer. Overall your answers would be:
a.) -5.42 m/s^2
b.) 3.8 m/s
Answered by
Nick
thanks
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