Question

urgent urgent... A motorcycle can be driven maximum with an acceleration of 6 m/s2, whereas its brakes can be applied to give maximum retardation of 12 m/s2. The minimum time, in which it can cover a distance of 800 metre, is
(1) 10 seconds (2) 20 seconds (3) 15 seconds (4) 40 seconds

Answer 1

Answer:

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Answer 2

Answer:

Given acceleration a= 5m/s

2

and maximum retardation =10 m/s

2

and distance between A and B= 1.5 km= 1500 m

We know acceleration a=

t

v−u

considering u= 0 m/s as initially the motorcycle is at rest.

So v=5t., where t is the time taken to accelerate.------(A)

As T is the total time than time taken for retardation by an amount of 10 m/s

2

be T-t.

From equation of motion we have retardation=

time

Finalvelocity−initial

In case of retardation or deceleration final velocity will be 0 as the motorcycle will come to rest.

So −v=−10(T−t)-------(B)

Putting the value of A in B we get

−5t=−10T+10t

T=

2

3t

-------(C)

As the total distance covered by the motorcycle in T seconds=1500 m,

thus

2

1

vT=1500

vT=3000

As earlier found v=5t and T=

2

3t

5t

2

3t

=3000

t

2

=400

t=20s

we know T=

2

3t

T=30s

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