Question

43. A car moving with a velocity of 30ms-1 is subjected to uniform retardation of 3m/s2.
The time taken by the car to come to rest is
5sec
10 sec
15 sec
20 sec​

Answer 1

Explanation:

Here,

u=30m/s,

a= -3m/s^2,

v=0.

To find: t

v=u+at

0=30+(-3t)

-3t=-30

t=-30/-3

t= 10 seconds

10 seconds is your answer.

Answer 2

Solution :

$\dag \: \green{ \underline{ \underline{ \sf{ \bold{Given :}}}}} \\ \\ \dashrightarrow \sf \: Initial \: velocity = 30mps \\ \dashrightarrow \sf \: Final \: velocity = 0 \\ \dashrightarrow \sf \: Retardation = 3m {s}^{ - 2} \\ \\ \dag \: \orange{ \underline{ \underline{ \sf{ \bold{ To \: Find :}}}}} \\ \\ \dashrightarrow \sf \: Time \: taken \: by \: car \: to \: stop \\ \\ \dag \: \red{ \underline{ \underline{ \bold{ \sf{Formula :}}}}} \\ \\ \dashrightarrow \sf \: As \: per \: first \: equation \: of \: kinematics... \\ \\ \gray{ \circ} \: \underline{ \boxed{ \bold{ \sf{ \pink{ \large{v = u + at}}}}}} \: \gray{ \circ}$$\dag \: \blue{ \underline{ \underline{ \bold{ \sf{ Terms \: Indication :}}}}} \\ \\ \divideontimes \sf \: v \: denotes \: final \: velocity \\ \divideontimes \sf \: u \: denotes \: initial \: velocity \\ \divideontimes \sf \: a \: denotes \: acceleration \\ \divideontimes \sf \: t \: denotes \: time \: interval \\ \\ \dag \: \purple{ \underline{ \underline{ \sf{ \bold{Calculation :}}}}} \\ \\ \rightarrowtail \sf \: 0 = 30 - 3t \\ \\ \circledast \sf \: \red{negative \: sign \: shows \: retardation} \\ \\ \rightarrowtail \sf \: 30 = 3t \\ \\ \rightarrowtail \sf \: t = \dfrac{30}{3} \\ \\ \rightarrowtail \: \underline{ \boxed{ \bold{ \sf{ \orange{ \large{time = 10 \: s}}}}}} \: \red{ \divideontimes}$