Math grade 9, can anyone solve it

a) Acceleration = 2.4 m per seconds²

b) Distance traveled = 132 m

Step-by-step explanation:

a). Acceleration of an object is defined by the equation,

v = u + at

Here, v = Final speed

u = Initial speed

a = Acceleration

t = Duration Or time

From the graph attached,

Between t = 0 and t = 5 seconds,

v = 12 meter per second

u = 0

t = 5 seconds

B substituting values in the equation,

12 = 0 + a(5)

a =

a = 2.4 m per seconds²

b). Equation representing distance traveled by an object,

v² = u² + 2as

Here, v = Final speed

u = Initial speed

a = Acceleration

s = Distance traveled

By substituting these values in the given equation,

1). Between t = 0 to t = 5 seconds,

(12)² = 0 + 2(2.4)s

s =

s = 30 m

2). Between t = 5 to t = 12 sec

Car moved with a constant speed (Acceleration = 0)

Distance traveled = Speed × Time

                             = 12 × 7

                             = 84 m

3). Acceleration of the car between t = 12 sec to t = 15 sec

By substituting the values in the equation,

v = u - at

0 = 12 - a(3)

a = 4 meter per seconds²

Distance traveled in this duration,

v² = u² - 2as

0 = (12)² - 2(4)s

144 = 8s

s =

s = 18 m

Therefore, total distance covered by the car = 30 + 84 + 18

= 132 meters

answer:  

123455421

step-by-step explanation: easy

114 plus 113 times 2 and 114 times 3