Question

Initial velocity of a car is
36 km/ur find the
distance
the after min,
if it goes with acceleration 2 mln
travelled by
choose the correct answere 4 x 1=4m_​

Answer 1

Correct Question :

Initial velocity of a car is  36 km/h . Find the  distance  after min,  if it goes with acceleration 2 m/s²

a. 4000 m

b. 4200 m

c. 4400 m

d. 4600 m

Solution :

Initial velocity , u = 36 km/h

$\to \sf u=36\times \dfrac{5}{18}\ m/s\\\\\to \sf u=10\ m/s$

Time , t = 1 min

• t = 60 s

Acceleration , a = 2 m/s²

Apply 2nd equation of motion ,

$\bigstar\ \; \sf s=ut+\dfrac{1}{2}at^2\\\\\to \sf s=(10)(60)+\dfrac{1}{2}(2)(60)^2\\\\\to \sf s=600+3600\\\\\leadsto \sf \purple{s=4200\ m}\ \; \bigstar$

Option b

Answer 2

Correct Question :

Initial velocity of a car is 36 km/hr. Find the distance after one min, if it goes with acceleration 2m/s².

[A] 4000 m.

[B] 4200 m.

[C] 4400 m.

[D] 4600 m.

Solution :

Given that,

• Initial Velocity = 36km/hr = 10m/s.

• Acceleration = 2m/s².

• Time = 1 min = 60 seconds.

We need to find,

• Distance traveled.

Formula Used,

$\sf {s = ut + \dfrac{1}{2} a {t}^{2} }$

Here,

s = Distance,

t = Time,

u = Initial Velocity,

a = Acceleration.

Apply the Formula.

$\to \: \sf{s = 10(60) + \dfrac{1}{\cancel{2}} \times \cancel{2} \times {(60)}^{2}}$

$\sf{\to \: s = 600 + 60 \times 60 }$

$\sf{\to \: s = 600 +3600 }$

$\underline{\boxed{\bf\red{\to s = 4200m. }}}$

Hence, Correct Answer = Option B.