Question

A cyclist starts from place A and covers a distance of 2400 to reach the place B.If the maximum possible acceleration of the cyclist is 2ms^-1 and its maximum retardation is 4ms^-1,what is the least time the cyclist can undertake the journey from A to B.

Answer 1

Answer:

this is the answer for question

Answer 2

Answer:

The least time the cyclist can undertake the journey from A to B is 60 seconds.

Explanation:

S₁+S₂=2400m     (1) (given in question )

For maximum acceleration

$S_1=\frac{1}{2}at^2=\frac{1}{2}\times 2\times t^2_1=t^2_1$       (2)

And the final velocity,

$v=at_1=2t_1$      (3)

For maximum retardation

The deacceleration given in the question is 4m/s2

$v=u+a_2t_2$     (4)

v=0 (as the cyclist is retarding)

u=2t₁

a₂=-4ms⁻²   (negative because the cyclist is retarding)

$0=2t_1-a_2t_2$

$a_2=\frac{2t_1}{t_2}$

$4=\frac{2t_1}{t_2}$

$t_2=\frac{1}{2}t_1$        (5)

Also,

$v^2=u^2+2a_2S_2$ (6)

$0^2=4t_1^2-2\times 4\times S_2$

$S_2=\frac{1}{2}t_1^2$        (7)

By substituting S₁ and S₂ in equation (1) we get;

$S_1+S_2=t^2_1+t^2_2=t^2_1+\frac{1}{2}t^2_1=2400$

$t^2_1=\frac{2400\times 2}{3}=1600$

$t_1=40sec$

$t_2=20sec$

Hence, the least time the cyclist can undertake the journey from A to B is 60 seconds.