i need help again.. coz i'll pass this tomorrow?

Question

A car starting from ret accelerates for 15.0 min until its velocity is 20 m/s. It then moves at constant velocity for another 20.0 min before it slows down and finally tops in another 10.0 min. Find (a) acceleration during the first 15 min, (b) the deceleration during the last 10min of its motion, and (c) total displacement

Answer 1

15 mins is 900 seconds.
20 mins is 1200 seconds
10 mins is 600 seconds.

you need to be consistent with your units.

a) acceleration = velocity change / time taken
therefore; a = 20/900 = 0.022 m/s^2

b) exactly the same thing, deceleration is the same as acceleration, just in a different direction (usually opposite).
therefore; a = 20/600 = 0.033 m/s^2

c) If you draw this motion on a graph, with velocity on the y axis, and time on the x, then the area under is equal to the total displacement (x = v*t). divide it into three areas, the two triangles on either end where the car accelerates and decelerates, and the rectangle in the middle where it has constant velocity.

acceleration triangle area;
(20 * 900) / 2 = 9,000 meters

constant velocity rectangle area;
20 * 1200 = 24,000 meters

deceleration triangle area;
(20 * 600) / 2 = 6,000 meters

Therefore the total distance traveled is 39 km.

Answer 2

V = A*T and A = V/T if starting from 0 velocty and having a constant acceleration

a. T = 15 min = 900 seconds
V = 20 m/s
A = 20/900 = 0.0222 m/s^2

b. Velocity is 20 m/s and the car stops in 10 min
V = V0 - A*T since there is an initial velocity and the minus sign is because of deceleration.

V = 0 (it stops) = 20 - A*T where T = 10*60 = 600 seconds
20 = A600 and A = 20/600 = 0.0333 m/s^2

c. displacement = D1 + D2 + D3

D1 is during acceleration.
D1 = (1/2)A*T^2 = (1/2)(0.2222)900^2 = 9000 m

D2 is during constant speed period. V= 20 m/s and T=20*60 = 1200 seconds
D2 = V*T= 20*1200 = 24000 m

D3 is during deceleration.
D1 = (1/2)A*T^2 = (1/2)(0.3333)600^2 = 6000 m

total displacement = 9000 + 24000 + 6000 = 39000 m