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.​

Answer 1

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.

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

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

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

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

Answer 2

Answer:

Given that,

Initial Velocity = 36km/hr = 10m/s.

Acceleration = 2m/s².

Time = 1 min = 60 seconds.

We need to find,

Distance traveled.

Formula Used,

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

2

1

at

2

Here,

s = Distance,

t = Time,

u = Initial Velocity,

a = Acceleration.

Apply the Formula.

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

2

1

×

2

×(60)

2

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

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

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