Question

. A car starting from rest and moving with a uniformacceleration attains a speed of 25 m/s in 5 seconds. Find a) The acceleration b)The distance travelled.

Answer 1

Given:-

• Initial Velocity of the car = 0m/s

• Final Velocity of the car = 25m/s

• Time = 5s.

To Find:-

• The Acceleration of the Car.

• Distance travelled by the Car.

Formulae used:-

• v = u + at

• S = ut + ½ × a × t²

Where,

v = Final Velocity.

u = Initial Velocity

a = Acceleration

t = Time

S = Distance.

Now,

→ v = u + at

→ 25 = 0 + a(5)

→ 25 = 5a

→ a = 25/5

→ a = 5m/s²

The Acceleration of the Car is 5m/s²

Therefore,

→ s = ut + ½ × a × t²

→ s = (0)(5) + ½ × 5 × (5)²

→ s = 0 + ½ × 125

→ s = 125/2

→ s = 62.5m

Hence, The Distance travelled by car is 62.5m

Answer 2

Given:-

•Initial velocity of the car(u)=0 (as it starts from rest)

•Final velocity of the car(v)=25m/s

•Time taken(t)=5 seconds.

To find:-

•Acceleration of the car

•Distance travelled by the car

Solution:-

By using the 1st equation of motion,we get:-

=>v=u+at

=>25=0+a×5

=>25=5a

=>a=25/5

=>a=5m/s²

Now,by using the 2nd equation of motion,we get:-

=>s=ut+1/2at²

=>s=0×5+1/2×5×(5)²

=>s=1/2×5×25

=>s=62.5m

Thus:-

•Acceleration of the car is 5m/s².

•Distance travelled by the car is 62.5m.

Some Extra Information:-

The 3 equations of motion,for a body moving with uniform acceleration are:-

• v=u+at

•s=ut+1/2at²

•v²-u²= 2as

Where:-

•v is final velocity of the body

•u is initial velocity of the body

•s is distance travelled by the body

•a is acceleration of the body

•t is time taken.