TS EAMCET 19-July-2022 Shift 1 Solved Paper

Initial velocity of motor car, u=7m∕ s
Stopping distance s=10m
Final velocity, v=0m∕ s
If retardation of car is am∕ s2, due to applying brakes, then
a=‌

v2−u2

2s

=‌

02−72

2×10

=‌

−4.9

2

m∕ s2
Negative sign indicates that retardation is applied in opposite direction of motion.
∴‌ Resistance, ‌R=ma=‌

w

9.8

×‌

−4.9

2

=−(‌

w

4

)