Question

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

Answer 1

t1 ≈ 10 sec

by using equation of motion, s = ut + (1/2) at²we get

100 = u x 10 + (1/2) a (10)² = 100

 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

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

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

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

= 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

= 10a = 5 = a = 5/10

= a = 1/2

substituting 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:

62.5m/s

Explanation:

$a= (x_{2} -x_{1})/(t_{2}/t_{1})$

a= 150-100 / 10-20

= 50/-20 = -2.5m/s^2

Δt = (10+10)s = 20 seconds

Δx= (100+150)m = 250 m

v= Δx/Δt , where x=displacement

= 250/20 = 12.5m/s

$v= u+at$

12.5 = u+(-2.5) X 20

12.5 = u-50

u = 50+12.5

$u= 62.5 ms^{-1}$

Therefore, u= 62.5 m/s

Please report to me any errors in the calculations