Question

Two cars A and B are moving towards each other on a straight road with velocity 10 m/s and 20 m/s respectively. When they are 150 m apart, then car A 5 decelerates with m/s2 and car B decelerates with 3 10 m/s2 until they stop in order to avoid collision. If 3 the distance (in m) between them when they come to rest is 15 x k, then find k



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Answer 1

Answer:

Distance between the two cars is 150 m.

For Car−1:

v=u

1

−at

0=10–2t

t = 5 sec

so distance travelled will be

s

1

=u

1

t−

2

1

at

2

s

1

=10×5−

2

1

×2×5

2

s

1

=25m

For Car−2:

v=u

2

−at

0=12–2t

t=6sec

so distance travelled will be

s

2

=u

2

t−

2

1

at

2

s

2

=12×6−

2

1

×2×6

2

s

2

=36m

So, when both cars will stop then the distance between them, will be :

150m–(25+36)m

Answer 2

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Given: Two cars A and B moving towards each other with velocity 10m/s and 20m/ s respectively. When they are 150m apart, A and B decelerate by 5 m/s^2 and 10m/ s^2 respectively.

To find: The constant k where 15k is the distance between them after coming to rest

Explanation: For car A, final velocity(v)= 0 initial velocity(u)= 10m/s, acceleration(a)= 5 m/s^2

Now, using kinematics formula, v= u- at

=> 10 = 5* t

=> t=2s Distance travelled in 2s = ut- 1/2at^2

= 10*2- 1/2*5*4

= 10 m

For car B, final velocity(v)= 0

initial velocity(u)= 20m/s, acceleration(a)= 10 m/s^2

Now, using kinematics formula, => u- at

=> 20 = 10* t

=> t=2s

Distance travelled in 2s = ut- 1/2at^2 = 20*2-

=1/2*10*4

= 20 m

Therefore, total distance covered by A and B=10+20= 30m

Distance between them when

to rest= 150-30= 120 m

they come

Now, 120 can be expressed in the form 15*8 which is same as 15*k.

Therefore, the value of k is 8.