HELP ME PLZ.****A cyclist starting from rest rolls down a 12meter ramp,when he gets to the bottom of ramp his?

Question

speed is 7.70m/s.

A.) Determine the magnitude of his acceleration ,assumed to be constant.

B.) If he ramp is inclined at 25degrees with respect to the ground,what is the component of his acceleration that is parallel to the ground?


I would gladly appreciate your help,thank you.

Answer 1

Velocity is a*t, so a = v/t; the distance traveled under constant acceleration is s = 0.5*a*t^2. t = √[2*s/a]. Put this into the first eq to get

a = v/√[2s/a], a^2 = v^2*a/2s a = v^2/2s a = 2.47 m/sec^2

The component of accel. parallel to the ground is a*cosø; a from above is 2.47, and cos(25º) = 0.906 so that component is 2.24 m/sec^2