A bike is moving in a straight line with a speed of 40km/h and behind the bike there is a car travelling with a velocity of 60 km/h in same direction . If car us given an accleration of 20km/h ^2 when the distance between them is 5km, then find the time in which the car can catch up with the bike?
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6
Let's make the bike stationary and give it's velocity to the car.
Now the relative velocity oftje car with respect to the car becomes
= 60-40
= 20 km/hr
given acc = 20km/hr²
Distance to cover = 5km
Using equation of motion
S= ut+(1/2)at²
5=20t +(1/2)20t²
10t²+20t-5=0
Using quadratic formula to solve for t
t =[-20+-√(20²-4(10)(-5))]/(2x10)
t=[-20+-√(400+200)]/20
t=-1 +-( 10√6/20)
t = (-1)+- (√6/2)
t =(-1) +- 1.22
t= 0.22 sec [ as negative value is not possible for time]
Now the relative velocity oftje car with respect to the car becomes
= 60-40
= 20 km/hr
given acc = 20km/hr²
Distance to cover = 5km
Using equation of motion
S= ut+(1/2)at²
5=20t +(1/2)20t²
10t²+20t-5=0
Using quadratic formula to solve for t
t =[-20+-√(20²-4(10)(-5))]/(2x10)
t=[-20+-√(400+200)]/20
t=-1 +-( 10√6/20)
t = (-1)+- (√6/2)
t =(-1) +- 1.22
t= 0.22 sec [ as negative value is not possible for time]
velanshyam4130:
answer is 13.24 min
Answered by
3
2.24 s is the right answer
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