A 1500 kg car moving at 5.0 m/s is initially traveling north in the positive y d

Question

A 1500 kg car moving at 5.0 m/s is initially traveling north in the positive y direction. After completing a 90° right-hand turn to the positive x direction in 4.6 s, the inattentive operator drives into a tree, which stops the car in 350 ms.

(a) In unit-vector notation, what is the impulse on the car during the turn?
(b) In unit-vector notation, what is the impulse on the car during the collision?

(c) What is the average force that acts on the car during the turn?

(d) What is the average force that acts on the car during the collision?

(e) What is the angle between the average force in (c) and the positive x direction?

Explanation / Answer

impulse =change in momentum a)Impulse1=1500 *5 i-1500*5 j=7500i-7500j b)impulse2=0 i-1500*5 i=-7500 i c)force1=7500 i-7500 j/4.6=1630.43(i-j) N d)force2=-7500 i/0.350=-21428.57 i N e)force1+force2=-19298.14 i-1630.43 j N angle=180+tan inverse(-1630.43/-19298.14)=184.8 from

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