Question

The velocity of a car travelling on a straight road is
36 kmh-1 at an instant of time. Now travelling with uniform
acceleration for 10 s, the velocity becomes exactly double. If
the wheel radius of the car is 25 cm, then which of the
following numbers is the closest to the number of revolutions
that the wheel makes during this 10 s. [WB-JEE 2013]​

Answer 1

Answer:

EXPLANATION :

GIVEN:

Initial velocity (u) =36 km/h =36×(5÷18) = 10 m/s

time(t) =10 sec

Final velocity (v) = 2u = 20 m/s

Radius of wheel(r) = 25cm = 0.25 m

Number of revolutions (N) = ???

Solution

As Object is moving with uniform acceleration so we can use Equations of Motion.

As,

$v = u + at$

=> a = (v-u)/t

=> a = 10/10

=> a = 1 m/s^2

Also,

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

=> Displacement (s) = 10(10) + (1/2 × 1 × 10 × 10)

=> s = 100 + 50

s = 150 m

So, Distance covered in One Revolution = 2* pi * r = 2×3.14×0.25

= 1.57 m

Now,

No. of Revolutions (N) = 150/1.57 = 95.5 Revolutions

Or We can say 96 Revolutions

Answer 2

Answer:

95 revolutions

Explanation:

velocity of car =10 m/s

time for which it accelerates=10 s

acceleration => v=u+at

final velocity is =20 m/s

acc=1 m/s^2

now radius of wheel =25 cm =1/4 m

in one round it covers distance = 2πr

=>π/2

now total distance covered in 10 seconds => s=ut+1/2at^2

s=150 m

now total revolutions=

150 /(π/2)

=>95.54

=95