Speeds of two identical cars are $U$ and $4 U$ at a specific instant. Assuming same retardation, the ratio of the respective distances in which the two cars are stopped from that instant is:

1 : 1

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

1 : 4

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

1 : 8

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

1 : 16

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

Open in App

Solution

The correct option is D $1 : 16$
Assuming same retardation:
Case 1:
$v^{2} = u^{2} + 2 a s,$
$v = 0, u = U, a = - a$
So, $s_{1} = \frac{U^{2}}{2 a}$
Case 2:
$v = 0, u = 4 U, a = - a$
So, $s_{2} = \frac{(4 U)^{2}}{2 a} = \frac{16 U^{2}}{2 a}$
So, $s_{1} : s_{2} = 1 : 16$

Suggest Corrections

Similar questions

Q. Speeds of two identical cars are

u

and

4 u

at a specific instant. If the same deceleration is applied on both the cars, the ratio of the respective distances in which the two cars are stopped from that instant is

Speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distances at which the two cars are stopped from that instant is

Q. Speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distances at which the two cars are stopped at the same instant is

Q. Speeds of two identical cars are

u

and

4 u

at a specific instant. If the same deceleration is applied on both the cars, the ratio of the respective distances in which the two cars are stopped from that instant is

Q. Speeds of two identical cars are u and 4u at a sufficient instant. What is the ratio of the respective distance at which the two cars are stopped from that instant?

Join BYJU'S Learning Program

Speeds of two identical cars are U and 4U at a specific instant. Assuming same retardation, the ratio of the respective distances in which the two cars are stopped from that instant is:

The correct option is D 1:16Assuming same retardation:Case 1:v2=u2+2as,v=0,u=U,a=−aSo, s1=U22aCase 2:v=0,u=4U,a=−aSo, s2=(4U)22a=16U22aSo, s1:s2=1:16

Speeds of two identical cars are $U$ and $4 U$ at a specific instant. Assuming same retardation, the ratio of the respective distances in which the two cars are stopped from that instant is:

The correct option is D $1 : 16$
Assuming same retardation:
Case 1:
$v^{2} = u^{2} + 2 a s,$
$v = 0, u = U, a = - a$
So, $s_{1} = \frac{U^{2}}{2 a}$
Case 2:
$v = 0, u = 4 U, a = - a$
So, $s_{2} = \frac{(4 U)^{2}}{2 a} = \frac{16 U^{2}}{2 a}$
So, $s_{1} : s_{2} = 1 : 16$