Question

C
2. A car is moving at 4 ms. After 5 seconds it has speeded up to 19 ms-1.

a.What is the acceleration of the car


b.how far deos the car travel during those 5s​

Answer 1

Answer:

accelaration of car is (v-u)/t  where v is final velocity and u is initial velocity and t is time .  Therefore  ( 19-5) / 4 i.e. $3.5$$m/s^{2}$ is the accelaration.

and for distance the formula is $(v^{2} -u^{2} )/2a$  where a is accelaration (3.5 calculated above therefore the answer is 96 m .

Answer 2

Answer:

3ms-2 and 56 m

Explanation:

acceleration(a) = (final velocity - initial velocity) /time taken

initial velocity (u) = 4ms-1

final velocity ( v) = 19ms-1

time taken (t) = 5 seconds

acceleration = ( v-u) /t

= (19-4)/5

= 15/5

= 3ms-2

from third equation of motion, we have

v² - u² = 2as

(19)² - (5)² = 2*3*s

(19+5) (19-5) = 6s from a² - b² = (a+b) (a-b)

24 *14 = 6s

336 = 6s

s = 336/6

= 56m ans

hope u got it

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