Question

Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far does it go in this time?

Answer 1

Answer:

(A)  $a=2.0.37m/sec^2$

(B) s = 146.664 m

Explanation:

We have given car starts from the rest so initial velocity u = 0 m /sec

Final velocity v = 88 km/hr

We know that 1 km = 1000 m

And 1 hour = 3600 sec

So $88km/hr=88\times \frac{1000}{3600}=24.444m/sec$

Time is given t = 12 sec

(A) From first equation of motion v = u+at

So $24.444=0+a\times 12$

$a=2.0.37m/sec^2$

So acceleration of the car will be $a=2.0.37m/sec^2$

(b) From third equation of motion $v^2=u^2+2as$

So $24.444^2=0^2+2\times 2.037\times s$

s = 146.664 m

Distance traveled by the car in this interval will be 146.664 m