Question

A body covers 18 m in 2 seconds and 34 m in next 2 seconds when moving in straight line having uniformly accelerated motion. Find the magnitude of acceleration involved.
$1) \: 2m {s}^{ - 2}$
$2) \: 4m {s}^{ - 2}$
$3) \: 5m {s}^{ - 2}$
$4) \: 10m {s}^{ - 2}$
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Answer 1

Answer:

Looking at the first case,

We have,

s = 18 m

t = 2 sec

By using Second Equation Of Motion,

s = ut + 1/2at^2

→18 = 2u + 2a

→18 = 2(u+a)

→u+a = 9. ----------------------------------(1)

Now, In the second case,

s = 18 + 34 = 52 m (total distance)

t = 2 + 2 = 4(total time)

By using Second Equation Of Motion,

s = ut + 1/2at^2

→52 = 4u + 16/2a

→52 = 4u + 8a

→52 = 4(u+2a)

→u + 2a = 13 ----------------------------------(2)

Subtracting (1) from (2),

u + 2a = 13

- u + a. = 9

----------------------------

a = 4 m/s^2

Therefore, the correct option will be 2.

Answer 2

Let initial velocity is u and uniform acceleration is a.

case 1 : body covers 18m in first 2 sec

so, s = 18 m , t = 2 sec

now using formula, s = ut + 1/2 at²

⇒18 = u(2) + 1/2a(2)²

⇒18 = 2u + 2a

⇒u + a= 9 .......(1)

velocity after 2 sec, v = u + a(2) = u + 2a

case 2 : body covers 34 m next 2 sec

so, initial velocity , u' = v = u + 2a

s' = 34m , t' = 2

so, s' = u't' + 1/2 a't'²

⇒34 = (u + 2a)(2) + 1/2 a(2)²

⇒34 = 2u + 4a + 2a

⇒17 = u + 3a ........(2)

solving equations (1) and (2),

17 - 9 = u + 3a - (u + a) = 2a

⇒a = 4

hence, uniform acceleration of body is 4m/s²