Given S = 3t + 6t? (in SI units) then
a) the initial velocity is 3 ms-1
b) the acceleration is 12 ms-2
c) the displacement of the particle after 2s is 30 m d) the displacement of the particle after 4 s is 30 m
Answers
Explanation:
Car X starts at 5 pm at a speed of 60 km/hr. Another car Y starts from the same point at 6:20 pm at a speed of 80 km/hr. At what time will X and Y meet each other?
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(Assuming the speed to be constant throughout the journey)
Speed of X : 1 km/min
Speed of Y : 1.33 km/min
Thus when Y start her/his journey (at 6.20 pm), distance travelled by X (who started at 5.00 pm);
5.00 pm to 6.20 pm = 80 minutes
Additional distance covered by X = (1 km/min)*(80 min) = 80 kms.
Thus, when Y started; distance between the two was 80 kms.
After 1 hr. (7.20 pm) distance-gap will be 60 kms. (speed difference)
After 2 hr. (8.20 pm) distance-gap will be 40 kms.
After 3 hr. (9.20 pm) distance-gap will be 20 kms.
After 4 hr. (10.20 pm) distance-gap will be 0 km. (first meet)
Hence at 10.20pm (4 hrs. after Y started journey), X and Y will meet for first time