Here is the solution and explanation for KEAM 2015 Mathematics Version B4 Question 5. For more answers and news, follow us on Facebook.

KEAM 2015 Mathematics Version B4 Question 5

If s = 2t3-6t2+at+5 is the distance traveled by a particle at time t and if the velocity is -3 when its acceleration is zero, then the value of a is : 

(A) -3            (B) 3             (C) 4             (D) -4              (E) 2

Answer

(B) 3  

Explanation

We know, from basic physics, that velocity is the first derivative of distance (or displacement) and acceleration is the first derivative of velocity (or second derivative of distance traveled). So, we have,

v = ds/dt and a = dv/dt = d2s/dt2

In our case,

v = 6t2-12t+a

a = 12t-12

It is given that, velocity v=-3 when acceleration a=0

So, a = 12t-12 = 0

i.e. t = 1

Substituting this value of t in v and equating v to -3, we can find a.

v(t) = 6t2-12t+a

We know that, v(1) = -3

Therefore, -6+a = -3

i.e. a = 3

So, our answer is (B) 3