InterviewSolution
Saved Bookmarks
| 1. |
A bus starts from rest with a constant accelerationof 5 ms . At the same time a car travelling with aconstant velocity of 50 ms overtakes and passesthe bus. (i) Find at what distance will the busovertake the car ? (i) How fast will the bus betravelling then ? [Ans. () 1000 m (i) 100 ms1 |
|
Answer» DISTANCE COVERED BY CAR WILL EQUAL TO THE DISTANCE COVERED BY BUS. SO, UT + 1/2 AT² = 50T 0×T+1/2×5×T² = 50T 5/2 T² = 50T 5/2 T = 50 T = 50×2/5 T = 20 SECONDS DISTANCE = SPEED × TIME DISTANCE = 50 × 20 DISTANCE = 1000 METRES USING 3RD EQUATION OF UNIFORMLY ACCELERATED MOTION, V² = U² + 2AS V² = 0² + 2×5×1000 V² = 10×1000 V = √10000 V = 100 M/S |
|