Answer to Question #84257 in Mechanics | Relativity for juan carlo buenabajo

Question

an object moving at 17.75 m/s with constant accelaration covers a distance of 90 m in 17.8 sec. what will be its velocity after covering 90 m? how long will it take for the object to change it velocity to 32.47 m/s? how far will it take for the object to change its velocity to 3.3 m/s?

Answer 1

After covering 90 m the velocity will be (remember that the object accelerates uniformly):

$$v_1=v_0+at_1=v_0+2s/t_1=17.75+2\cdot 90/17.8=27.86 \text { m/s}.$$

Changing velocity from 27.86 m/s to 32.47 m/s with the same acceleration will take

$$t_2=\frac{v_2-v_1}{a}=\frac{v_2-v_1}{2s/t_1^2}=\frac{32.47-27.86}{2\cdot 90/17.8^2}=8.1 \text{ s}.$$

A distance for slowing down from 32.47 to 3.3 m/s (breaking distance) depends on deceleration

$$a_b$$

which is not mentioned in the condition. Assuming that the deceleration is equal to the maximum g-force a human can withstand (

$$9g$$

), the breaking distance will be

$$L=\frac{v_3^2-v_2^2}{2\cdot a_b}=\frac{3.3^2-32.47^2}{2\cdot (-9)\cdot 9.8}=5.9 \text{ m}.$$