Question

A motorbike running at 90 km/h slowed down to 18 km/h in 2.5sec . Calculate (i) Acceleration (ii)distance covered

Answer 1

Hii
Here motorbike in retardation so
initial velocity (u)= 90 km/h = 25 m/s
final velocity (v) = 18 km/h = 5 m/s
time (t) = 2.5 s
acceleration (a) = ?
distance (s) = ?
a = v - u / t
= 5 - 25 / 2.5
= - 20 / 2.5
a = - 8
for distance apply motions third equation
2 a s = v^2 - u ^2
2 × ( -8 ) × s = ( 5× 5) - ( 25 × 25)
-16 × s = 25 - 625
s = - 600/ - 16
s = 37.5
Hope it help you

Answer 2

We know that 1 km/ hr = $\frac{5}{18}\ m/s$

The initial velocity of the motor bike is given as u = 90 km/ hr.

                                                                                  = 25 m/s

The final velocity of the motor bike is given as v = 18 km/hr.

                                                                                = 5 m/s

The time taken by the motor bike t = 2.5 s.

Let a is the acceleration of the body.

From equation of kinematics we know that,

                                             v = u + at

                                          ⇒at = v -u

                                          ⇒ at = 5 m/s - 25 m/s

                                            ⇒2.5t  = - 20 m/s

                                            ⇒ $a=\frac{-20}{2.5}\ m/s^2$

                                            $=\ -8\ m/s^2$

Here, the negative sign is used due to the fact that velocity is decreasing.

Let s is the distance traveled by the body .

Again from equation of kinematics, we know that -

                                $v^2-u^2=2as$

                               ⇒ $s=\frac{v^2-u^2}{2a}$

                                         $=\frac{(5)^2-(25)^2}{2\times (-8)}\ m$

                                         $=\frac{25-625}{-16}\ m$

                                         $=37.5\ m$         [ans]