A car traveling initially at #+7.0# #m##/##s# accelerates uniformly at the rate of #+0.80# #m##/##s^2# for a distance of #245# #m#. What is its velocity at the end of the acceleration? What is its velocity after it accelerates for 125 m? 67 m?

Question

30905 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-01T12:31:36

#v_245 = 21 m/s#

#v_125 ~~15.78 m/s#

#v_67 ~~12.5 m/s#

Summarising the information given:

Initial velocity #v_1 = 7.0 m/s#

Acceleration #a = 0.80 m/s^2#

Distance travelled #s_1 = 245 m#

From one of the key kinematic equations

#v_245^2 = v_1^2 +2*a*s_1#

#:.v_245^2 = 7^2 +2*0.8*245#

#v_245 = sqrt(441) = 21 m/s#

#v_125 =sqrt(49+2*0.8*125) ~~15.78 m/s#

#v_67 = sqrt(49 + 2*0.8*67) ~~12.5 m/s#