Question #101055

A ball rolls down a hill with constant acceleration. After 2.0 seconds the ball has travelled 1.2 m, how far will it have travelled after 4.0 seconds?

Expert's answer

As per the given question,


Intial velocity is not given, so i am considering it zero,

hence applying the second law of motion,

s=ut+at22s= ut+\dfrac{at^2}{2}

1.2=0+a×2×221.2=0+\dfrac{a\times 2\times 2}{2}

a=1.22=0.6m/sec2a=\dfrac{1.2}{2}=0.6 m/sec^2

Now, again applying the second law of motion,

s1=ut+at222=0+0.6×4×42=4.8ms_1=ut+\dfrac{at_2^2}{2}=0+\dfrac{0.6\times 4\times 4}{2}=4.8 m

After the next 4 sec ball will travel

S3−S1=a(t22−t12)2=0.6×(36−4)2=9.6mS_3-S_1=\dfrac{a(t_2^2-t_1^2)}{2}=\dfrac{0.6\times (36-4)}{2}=9.6 m


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