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As soon as a car just starts from rest in a certain dercation, a scooter moveing with a uniform speed overtakes the car. Their velocity-time graph is shown in . Calculate . a. The difference between the distances travlled by the car and the scooter in `15 s`, b. The distance of car and scooter from the starting point at that instant. |
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Answer» a. The distance travekked by car in `15 `s `=Area` of `Delta OAC` `=(1)/(2) xx 15 xx 45 =337. m` Distance travelled by scooter in `15 s` `=Area` of retangle `OCEF` `=15xx30=450 m` Thus, difference between distance travelled by them `=450 m-337.5 m= 112.5 m` b. Let aftre time `t` from start car will catch up the scooter. In time `t`, the distance travelled by them are equal. Distance travelled by car `=(1)/(2)xx 1545+ 45(t-15)` Distance travelled by scooter `=30t` `(1)/(2) xx 15 xx 45 +45 (t-15)=30t` which gives `t=22.5 s` Distance travelled by car or scooter in `22.5 s=30xx22.5` `=675 m` So the catr catches the scoter when both are at `67.5 m` from the starting point. |
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