InterviewSolution
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InterviewSolution
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A car travels with a uniform velocity of 20 m s^(-1) for 5 s. The brakes are then applied and the car is uniformly retarded. It comes to rest in further 8 s. Draw a graph of velocity against time. Use this graph to find : (i) the distance travelled in first 5 s, (ii) the distance travelled after the brakes are applied, (iii) total distance travelled, and (iv) acceleration during the first 5 s and last 8 s. |
Answer» Solution :The graph of VELOCITY against time is shown in FIG. 2.27.  m (i) The DISTANCE TRAVELLED in first 5 s = area of rectangle OABD = `OD xx OA = 5 s xx 20 m s^(-1) = 100 m` (ii) The distance travelled by car after the brakes are applied = area of `Delta BDC =(1)/(2)xxDCxxDB` `= (1)/(2) xx(13-5) s xx 20 m s^(-1) = 80 m`. (iii) Total distance travelled = area of rectangle OABD + area of triangle BDC = 100 + 80 = 180 m (iv) Acceleration in the first 5 s (in part AB) = 0 (since straight line AB is parallel to the time axis, so slope = 0). Acceleration in the last 8 s (in part BC) = Slope of the line BC `= (BD)/(DC) = ((0-20)m s^(-1))/((13-5)s)=(-20 m s^(-1))/(8 s )` `=- 2.5 m s^(-2)` Since acceleration is negative so retardation = `2.5 m s^(-2)` . |
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