Verified answer

Acceleration = (v/t) = 37/3.5, = - 10.57m/sec^2.

Acceleration to stop from 37m/sec in 6m. = -114.083m/sec^2.

I agree there is something screwy with the question.

You won't get ME in a car that stops from (37 x 3600) = 133kph. in just 6 metres! That's deadly!

Stopping in 3.5s. is more reasonable. Stopping distance = 1/2 (t^2 x a) = 129.48 metres.

I don't CARE whar Kushagra's mathematics says to make an answer agree with a choice!

That's nutty! Is that why you put "breaks to a stop"? It certainly would....

Distance to stop from 37m/sec at -20m/sec^2 = (v^2/2a) = 34.225 metres, not 6 metres.

The answer is correct. It is (B)

However, the car does not stop in 3.5 sec ----> This is wrong in question

==============================================================

Initial Velocity [u] = 37 m/s

Distance [h] = 6 m

Time [t] = 3.5 sec

h = ut + 1/2at^2 ---> use IInd Equation of motion

6 = 37*3.5 + 1/2(a)(3.5)^2 ----> Putting respective values

So, 6 = 129.5 + 6.125a => a = -123.25 / 6.125

So, acceleration = -20.1633 m/s^2 = -20 m/s^2

=======================================================================

Note : The sign of acceleration is negative because it is in the direction opposite to the direction of motion of car. The acceleration is stopping the car and not moving it forward.

acceleration a = v - u / t = 0 - 37 / 3.5 = - 10.6 m/sec^2

also a = 0 - u^2 / 2 s = - 37x37 / 2x 6 = - 114 m/sec^2

your given values are giving two different values of acceleration and these are not found in your options. i feel there is something terribly wrong in your question . look into it.

try wish123 another webiste that might help. it's a good question.

How come it broke?