car is travelling with a speed of 36km/h. The driver applies the brake and retards the car uniformly. The car is stopped in 10s. Find(1) retardation of the car. (2) distance travelled before it is stopped after applying the brakes.
Answers
Answer:
the retardation is -1 m/s^2
Explanation:
how to do it??
the answer is here......
during a retardation the final velocity is less than initial velocity........therefore as the car is stopping then the final velocity should be 0m/s
and the initial velocity should be 36 km/hr.......thus we have to change it to m/s........to change it we will multiply 1000 with 36 and divide it with 3600.......thus the answer comes 10 m/s..... now it is clear that you have to use the formula......
V=U+AT
(FINAL VELOCITY EQUALS INITIAL VELOCITY PLUS RETARDATION INTO TIME)
THEN......JUST PUT THE VALUES IN THEIR PLACE.........THE ANSWER COMES....
0=10+ A×10
THE ANSWER COMES
-1M/S^2
(minus sign proves that it is retardation)
now to take out the distance......
u have to use the formula.....
S =UT +1/2 A ×T ^2
SIMPLY PUT THE VALUES......
S= 10 ×10 +1/2 ×(-1)× 10^2
therefore the distance comes......(100 -50)
= 50 m
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