Question

Acceleration is applied 4m/s^2 to a car moving at a speed of 54 km / h for 5 sec. Find final velocity and distance


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Answer 1

V= U+at , U= 54km/hr( 54×5/18 m/sec = 15m/sec), a= 4m/sec2, t= 5 , V= 15+ 4×5 , V= 15+20, V= 35 m/sec , S= Ut+1/2 at^2, S= 15×5+ 4×5×5/2,. S= 75+50, S= 125 m

Answer 2

Given :-

• Initial velocity, u = 54 km/h

Converting 54 km/h to m/s :-

( Divide the value of velocity by 3.6)

Hence, initial velocity (u) = 15 m/s

• Acceleration, a = 4 m/s²

• Time taken, t = 5 sec

To Find :-

• Final velocity

• Distance covered by the car

Solution :-

To find final velocity, we have to use 1st equation of motion.

We know,

v = u + at

Where,

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

Now, put the given values

v = u + at

⟹ v = 15 + 4 × 5

⟹ v = 35 m/s

Hence, final velocity of the car is = 35 m/s

Again :-

To find distance ( s) , we we have to use 2nd equation of motion.

We know ,

$\pink{\boxed{\boxed{\rm S=ut+\dfrac{1}{2}at^{2}}}}$

Where,

s = Distance covered

By substituting the given values, we get

⟹ s = 15 × 5 + ½ × 4 × 5²

⟹ s = 125 m

Hence, distance covered by the car is = 125 m

✰✰Equations of motion ✰✰:-

• v = u + at (First equation of motion)

• s = ut + ½ at² (Second equation of motion)

• v² = u² + 2as (third equation of motion)