can someone help me with this question is really confusing and I don't understand, please.... I'm desperate
a ball along a straight path with the following velocity and time intervals:
-3.0s at 15.0m/s
-7.0s at 20.0m/s
-6.0s at -30.0m/s
-4.0s at 3.5m/s
a.total displacement?
b.displacement in the first 10s?
c.displcement in the last 10s?
d.displacement in the 5.0 - 15.0s time interval?
can you please show the procedure so I can understand how to do this equation. thank you
3 years ago
Answered By Majid B
a)
$d=\sum\left(v_i\times t_i\right)=\left(15\left(\frac{m}{s}\right)\times3\left(s\right)\right)+\left(20\left(\frac{m}{s}\right)\times7\left(s\right)\right)+\left(-30\left(\frac{m}{s}\right)\times6\left(s\right)\right)+\left(3.5\left(\frac{m}{s}\right)\times4\left(s\right)\right)=19m$d=∑(vi×ti)=(15(ms )×3(s))+(20(ms )×7(s))+(−30(ms )×6(s))+(3.5(ms )×4(s))=19m
b)
$d=\sum\left(v_i\times t_i\right)=\left(15\left(\frac{m}{s}\right)\times3\left(s\right)\right)+\left(20\left(\frac{m}{s}\right)\times7\left(s\right)\right)=185m$d=∑(vi×ti)=(15(ms )×3(s))+(20(ms )×7(s))=185m
c)
$d=\sum\left(v_i\times t_i\right)=\left(-30\left(\frac{m}{s}\right)\times6\left(s\right)\right)+\left(3.5\left(\frac{m}{s}\right)\times4\left(s\right)\right)=-166m$d=∑(vi×ti)=(−30(ms )×6(s))+(3.5(ms )×4(s))=−166m
d)
$d=\sum\left(v_i\times t_i\right)=\left(20\left(\frac{m}{s}\right)\times5\left(s\right)\right)+\left(-30\left(\frac{m}{s}\right)\times5\left(s\right)\right)=-50m$d=∑(vi×ti)=(20(ms )×5(s))+(−30(ms )×5(s))=−50m