2(a) A car accelerates from 22ms^-1 to 43ms*-1 in 18.6 seconds.
(a) Calculate the acceleration of the car.
2(b)
Find the distance covered by the car in 10 seconds.

2(c) Find the velocity after 5 seconds.

2a.

a=1.13ms^-2

2b.

S=277m

2c.

V=27.7ms-¹

Explanation:

Initial Velocity (U)=22m/s¹

Final Velocity (V)=43m/s²

Time(t) =18.6s

a. a=V-U/t

a=43-22/18.6

a=1.129

a=1.13m/s²

2b.

S=ut+1/2 at²

s=22(10)+1/2×1.13(10)²

s= 220+0.57(10)²

s= 220+0.57(100)

s= 220+57

s=277m

2c.

V=U+AT

V=22+1.13(5)

V=22+5.65

V=22+5.7

V=27.7m/s¹

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

a cold ice-tea.

pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015