Question

A car starts from rest and accelerates with 2 m/s

2 for 10 s. After maintaining the

velocity for 10 seconds, it comes to rest decelerating by 1 m/s

2

. Find the total distance

travelled.​

Answer 1

Answer:

500 metres

Explanation:

Given:

Initial velocity = u = 0 m/s (As the car starts from rest)

Acceleration = a = 2 m/s²

Time = t = 10 seconds

First we will find the distance in the acceleration period and then find the distance during decceleration period and add them both

Using first equation of motion,

v=u+at

v=0+2×10

v=20 m/s

Now using third equation of motion:

v²-u²=2as

20²-0²=2×2×s

400=4s

s=100 metres

distance during constant velocity = 20×10 = 200 metres

During decceleration:

Initial velocity = u = 20 m/s

Final velocity = v = 0 m/s

Acceleration = a = -1 m/s²

Using first equation of motion,

V=u+at

0=20-1t

t = 20 seconds

Now using third equation of motion:

v²-u²=2as

0²-20²=2×-1×s

-400= -2s

s = 200 metres

Total distance = 200+200+100 = 500 metres

The total distance covered is equal to 500 metres

Answer 2

Answer:

500m

Explanation:

Maximum velocity attained is 20 metres/second

Total distance travelled is 500 metres

Initial speed (U) = 0

Acceleration (A) = 2

Time (T) = 10

Final speed (V) = U + AT

                        = 0 + 10 x 2

                        = 20

Hence maximum velocity attained is 20 m/s

Deceleration (D) = -1

Initial velocity after 20 seconds = V = 20

Final velocity (V1) = 0

Time (T1) = (V1 - U) / D

               = (0 - 20) / (-1)

               = 20 seconds

Total distance travelled = Distance while accelerating + Distance while decelerating + Distance without acceleration

= (0.5 x 2 x 10^2) + (20^2 / 2 x 1) + (20 x 10)

= 100 + 200 + 200

= 500 metres