1) Speed of two identical cars are 'u' and '4u' at a specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is:
(A) 1:1
(B 1:4
(C) 1:8
(D) 1:16
2) Two particles start moving along the same straight line at the same moment from the same point. The first moves with constant velocity 'u' and the second with constant acceleration f. During the time that elapses before second catches the first, the greatest distance between the particles is:
(A) u/f
(B) u^2/2f
(C) f/2u^2
(D) u^2/f
Any answer will be appreciated
(A) 1:1
(B 1:4
(C) 1:8
(D) 1:16
2) Two particles start moving along the same straight line at the same moment from the same point. The first moves with constant velocity 'u' and the second with constant acceleration f. During the time that elapses before second catches the first, the greatest distance between the particles is:
(A) u/f
(B) u^2/2f
(C) f/2u^2
(D) u^2/f
Any answer will be appreciated
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1- Considering the retardation to be same as (-a) ..as uncle forgot to mention that in a hurry..
applying v^2 - u^2 = 2as ..
u^2 = 2as1
16u^2 = 2as2
s1/s2 = 1/16.
D) 1:16
2- Distance travelled by 1st particle up to time t = ut
Distance travelled by 2nd particle up to time t =(1/2)ft^2
Gap between them, s = ut - (1/2)ft^2
this gap will be maximum when ds/dt = 0 => u - ft = 0 => t = u / f
d2s/dt^2 = -f <0 , so t = u /f is a point of maxima
smax = u(u/f) - (1/2)(f)(u/f)^2 =
u^2 / 2f
B) u^2 / 2f
applying v^2 - u^2 = 2as ..
u^2 = 2as1
16u^2 = 2as2
s1/s2 = 1/16.
D) 1:16
2- Distance travelled by 1st particle up to time t = ut
Distance travelled by 2nd particle up to time t =(1/2)ft^2
Gap between them, s = ut - (1/2)ft^2
this gap will be maximum when ds/dt = 0 => u - ft = 0 => t = u / f
d2s/dt^2 = -f <0 , so t = u /f is a point of maxima
smax = u(u/f) - (1/2)(f)(u/f)^2 =
u^2 / 2f
B) u^2 / 2f
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1) Kinetic energy must be used to stop the vehicles. Since the KE = 1/2*m*v^2, the vehicle with 4 times the velocity will take 16 times as far to stop. answer D.
2) Answer B is correct because the dimensional analysis narrows the answer to B or D. The constant velocity particle and the accelerating particle is given by s = u*t - 1/2*f*t^2
Setting the derivative of s with respect to t at 0 = u - (1/2)*2*f*t t = u/f
2) Answer B is correct because the dimensional analysis narrows the answer to B or D. The constant velocity particle and the accelerating particle is given by s = u*t - 1/2*f*t^2
Setting the derivative of s with respect to t at 0 = u - (1/2)*2*f*t t = u/f
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1st answer is 1:4 taking time to be t in both cases as it is given specific instant and Instant means the same time.. so we know distance=speed x time so distance1=speed1 x time1 ratio distance2=speed2 x time2.. putting values distance1=1u x t and distance 2=4u x t taking D1 : D2 = 1 ut/4ut=1:4.. and in second question can u tel me the meaning of '^'?? Thanx