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:

$u = 36km {h}^{ - 1} = 36 \times \frac{5}{18} = 10 \\ t = 1min = 60s \\ a = 2m {s}^{ - 2} \\ s = ut + \frac{1}{2} a {t}^{2} \\ = 10 \times 60 + \frac{1}{2} \times 2 \times {60}^{2} \\ = 600 + 3600 \\ = 4200$

thus the distance traveled by the car is 4200m[B]