Recent content by S313

Could it be that since the slope is upwards, the acceleration has to have opposite sign of v0? if we choose the + x-axis to be upwards along the slope, then v0 will be positive, while acceleration will slow down and be negative (opposite to v0)? If so, then s= 3(1.2) + (1/2)(-2.5)(1.2)^2 = 3.6 -...

but if v0 = + 3 m/s, then: a = 3/1.2 = + 2.5 m/s^2 and s = v0t + 1/2at^2 = 3(1.2) + (1/2)(2.5)1.2^2 = 3.6 + 1.8 = 5.4 m

Yes: s = v0t + 1/2at^2 = (-3)(1.2) + (1/2)(-2.5)(1.2)^2 = -3.6 - 1.8 = - 5.4 m, but this is not the answer. The answer is 1.8 m according to the book.

a = -3/1.2 = -2.5 m/s^2, but how will this lead to finding how far up the slope it went?

Thanks BvU! Happy to be here! Even if we assume that v0 = -3 m/s, the acceleration will be: a = -v0/t = -3/1.2 = -3.6. Plugging this into s = v0t + 1/2at^2 will yield: -3.6 - 1.8 = -5.4 m. I won't get only 1.8 m. To get the final answer to be 1.8 m, I should assume v0=0 (then v0t = 0) and the...

Homework Statement The initial speed of a ball up an inclined plane is 3 m/s upwards. The ball reaches its highest point after 1.2 seconds. a) find the acceleration. b) how far has the ball gotten during this time? c) how far has the ball rolled after 0.4 s and after 2.4 seconds? Homework...