Question

Motion with constant acceleration
3. A car accelerates with uniform rate from rest on a straight road. The distance travelled in the last
second of a three second interval from the start is 15 m then find the distance travelled in first second
in m. / solve it plz​

Answer 1

Answer:

Let,

Point where the car is at rest = S,

Point where velocity = 5 m/s = A

Point where velocity = 10 m/s = B

Point where velocity = 15 m/s = C

We have :

At S, t = 0 sec

At A, t = 1 sec

At B, t = 2 sec

At C, t = 3 sec

We have following formula :

d=1/2(Vf+Vi)×t

Therefore,

SA=1/2∗(5+0)∗1=2.5

AB=1/2∗(10+5)∗1=7.5

BC=1/2∗(15+10)∗1=12.5

The total distance covered by body = SA + AB + BC

= 2.5 + 7.5 + 12.5

= 22.5 meters.

Explanation: