Question

A body moving with uniform acceleration covers 100m in first 10 seconds and 150 m in next 10 seconds what is the initial velocity of the body

Answer 1

Let initial velocity =u m/s, acceleration = a m/s² final velocity ( the velocity at the end of first 10 seconds) = v m/s

Now for fist case time t = 10 sec.

Using second equation of motion s = ut + (1/2) at² we get

100 = u x 10 + (1/2) a (10)² = 100 = u x 10 + (1/2 ) x a x 10 x 10

= 100 = 10{u + (1/2) x a x 10) = 100/10 = u + 5a ( rearranging equation)

= u + 5a = 10 .........(1)

Now for second case there is one important point to consider that the initial velocity for the second case will be the final velocity of the first case as this motion starts after the end of first motion.

Thus to find initial velocity for second case we use first equation of motion v = u + at on first motion we get

v = u + a x 10 = v = u + 10a

Now this will be our initial velocity for second motion

Using second equation of motion s = ut + (1/2) gt² for second motion we get

150 = (u + 10a)x10 + (1/2) a (10)² = 150 = 10x(u + 10a) + (1/2) g x10x10

= 150 = 10 {(u +10a) + (1/2) x 10 x a} = 150/10 = u + 10a + 5a (rearranging equation)

= 15 = u + 15a= u + 15a = 15 ......(2)

Now subtracting equation (1) from equation (2) we get

u + 15a - (u + 5a) = 15 - 10 = u + 15a - u - 5a = 5

= u - u + 15a - 5a = 5 (rearranging equation)

= 10a = 5 = a = 5/10

= a = 1/2

Putting this value of a in equation (1) we get

u + 5 x (1/2) = 10 = u + (5/2) = 10

= u = 10 - 5/2 = 10 - 2.5

= u = 7.5 m/s

Answer 2

Answer:

Step-by-step explanation:

From the second law of Newton’s eqn.,

100 = 10u+50a;

50 = 10u+ 40a;

50 = 10a;

a = 5 m/sec2;

150 = 10 u;

u = 15 m/sec