An automobile travelling with a speed of $60 k m / h$ , can brake to stop within a distance of $20 m$ . If the car is going twice as fast, i.e. $120 k m / h$ , the stopping distance will be:

20 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

40 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

80 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C $80 m$
as we know that $V^{2} = U^{2} + 2 a S$
so convert $(K m / h r) i n t o (m / s e c)$ we find $V_{1} = 1000 m / s e c$ and $V_{2} = 2000 m / s e c$ $S_{1} = 20 m$
now we get the following relation
$\frac{V_{1}^{2}}{S_{1}} = \frac{V_{2}^{2}}{S_{2}}$
put the value in above equation we will find $S_{2} = 80 m$

Suggest Corrections

Similar questions

60 k m / h

20 m

120 k m / h

60 {k m h}^{- 1}

20 m

120 {k m h}^{- 1}

60

20

120

60

20

Join BYJU'S Learning Program