Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

# A car accelerates from rest at 2 m/s² for 5 seconds, travels at constant speed for 10 seconds and....?

Cont...

decelerates to rest at 2 m/s². Calculate the distance travelled by the car.

Relevance

Let's divide the distances part by part

d1 = initial velocity x time + 1/2 x acceleration x time^2

= 0 x 5 + 1/2 x 2 x 5^2

= 1 x 25

25 m

velocity  = initial velocity + acceleration x time

d2 = velocity x  time

= (initial velocity + acceleration x time) x  time

=  0 + 2 x 5 x 10

=10x10

= 100m

deceleration = (initial velocity - final velocity) / time

2 = (10-0)/time

time = 10/2 = 5s

d3 = (initial velocity + final velocity)/2 x time

= (10 + 0)/2 x 5

= 10/2 x 5

= 5 x 5

= 25m

d1 + d2 + d3

=25 + 100 + 25

100+25+25

• Login to reply the answers
• Answer to this problem = 150 m

Time-Velocity Diagram of the movement is as follows.

From A to B, it accelerates @ 2 m/s²

Hence Tan θ = 2 m/ s² = BE/AE. But AE = 5 second

Hence BE = 10 m/s

From B to C it moves with a constant velocity = BE = 10 m/s

From C to D it moves with a retardation of 2 m/ s²

Hence slope of AB and CD are numerically equal to each other.

Hence Total distance ( D ) traveled = Area of the Trapezium ABCDA

=> D =  (1/2) ( AD + BC)*(BE)

D = (1/2) (20+10)*10 = 150 m ……………….. Answer • Login to reply the answers