a bus starts to move with an acceleration of 1m/s² .A man who is 48m behind the bus runs with constant velocity of 10m/s to catch it .In how much time he will catch the bus?
pls answer quickly with step by step explaination
Answers
Answer:
Answer is 8.7 seconds...
Answer:
Given:
Acceleration of the bus = 1m/s^2
Distance of the man = 48 meters
For solving this question we have to frame an quadratic equation:
The equation must be framed in the following way:
= Distance Covered by the Bus = Distance Covered by the man - 48
Taking these values in terms of physics:
1/2 a t^2 = vt - 48
Substituting all the values known to us in this formula:
1/2(1)t^2 = 10t - 48
Multiplying the Equation by 2 to simplify:
t^2 = 20t -96
Taking the LHS to the RHS:
t^2 - 20t + 96
We get a quadratic equation now solving this equation by splitting the middle term:
t^2 - 12t - 8t + 96
Taking t and - 8 common we get:
t (t - 12) - 8(t -12)
(t - 12) and (t -8)
t = 12 sec and t = 8 sec
Therefore the time taken can either be 12 seconds or 8 seconds but here in this question they asked the minimum time after which the man will catch the bus so this would be the less time of the two so the answer is 8 seconds.