Physics question?

Question

Norma releases a bowling ball from rest; it rolls down a ramp with constant acceleration. After half a second it has traveled 0.75 m. How far has it traveled after two seconds?
a.1.2 m
b.4.7 m
c.9.0 m
d.12 m

Answer 1

This problem is very interesting.

First, you have to find the unknown acceleration, then you can find the distance.

s = ut + 1/2at^2

0.75 = 0.5a * 0.5^2
a = 6 m /s^2

Now reapply s= ut + 1/2at^2 with the new t = 2s
s = 12m

So the answer is d. 12m.

Answer 2

Is it still on the ramp? If so, try this.

Use s = (1/2)*a*t^2 to get the acceleration

Use s = (1/2)*a*t^2 again with t = 2 sec

Answer 3

We know that : s = ut +1/2 at^

s=distance;u=initial velocity;a=acceleration;t=time;

given

u=0

=> 0.75= 1/2 a (1/2)^

=> a = 6.0 m/s^

now , we get :

s = 1/2 (6)(2)^

=> s = 12m . (ans) (d)

Answer 4

using s=u*t + 1/2 a* t* t;
.75=0 + .5 * a* .5 *.5
=>a= .75/.5*.5*.5=30/5=6

so after 2 secs
s=.5 * 6 * 2 *2
=>1.2 m

Answer 5

In four times the time, it will have traveled sixteen times the distance, hence d.

Answer 6

do your own homework