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

Step-by-step explanation:

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} }s=ut+

2

1

at

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}}→s=10(60)+

2

1

×

2

×(60)

2

2

1

×

2

×(60)

2

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

\sf{\to \: s = 600 +3600 }→s=600+3600

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