Question

A car is traveling at 70 km/h. It then uniformly decelerates to a complete stop in 12 s. Find its acceleration ( in m/s2 ).

Answer 1

Answer:

Acceleration will be $-1.620m/sec^2$

Explanation:

We have given initial speed of the car is 70 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $70km/hr=70\times \frac{1000m}{3600sec}=19.444m/sec$

It is given that car stops in 12 sec

So final speed of the car v = 0 m/sec

Time t = 12 sec

From first equation of motion v = u+at

So $0=19.444+a\times 12$

$a=\frac{-19.444}{12}=-1.620m/sec^2$ ( negative sign indicates that speed of the car will constantly decrease )

Answer 2

Answer:

- 1.62 m/s²

Explanation:

initial velocity, u = 70 km/h = 19.44 m/s

time, t = 12 s

final velocity, v  = 0

Acceleration is defined by the rate of change of velocity.

By use of first equation of motion

v = u + a t

where, a be the acceleration

0 = 19.44 + a x 12

a = - 1.62 m/s²

Thus, the acceleration is - 1.62 m/s².