Question

Mohan bought a new car and wanted to test it on highways. He thought he will find out
the acceleration of his car at different velocities in the first 6 seconds. He called his friend
Sham and told him to sit along side him and note down the different speeds. Sham
prepared the following graph. Mohan's son who studied in 9th class wanted to do an
experiment with the car. He had recently learned a peculiar thing about circular motion
and forced his father to take the car to circular track and drive at constant speed




a) find the cars acceleration from a to b
b) find out the cars acceleration from b to c
​

Answer 1

Answer:

(a) a = (25 - 0) / (3 - 0) = 8.3 m/s2

(b) a = 17 - 25 ) / (4 - 3) = -8 m/s2

(c) Distance = Area of triangle ABE = 1 / 2 × 3 × 25 = 37.5 m  

(d) V = (17 - 0) / 2 = 13.75 m/s

(e) This Distance = Area of trapezium BCFE  

= 1 / 2 (25 + 17) × (4 - 3) = 21m

please mark me as..

Answer 2

a) Acceleration from A to B is 8.33 $m/s^{2}$

b) Acceleration from B to C is -8 $m/s^{2}$

Given,

According to the given graph we have,

initial velocity in the first 3 seconds (v1) = 0 m/s

final velocity in the first 3 seconds (v2)= 25 m/s

initial velocity from 3 to 4 seconds (v2)= 25 m/s

final velocity from 3 to 4 seconds (v3)= 17 m/s

To Find,

a) Acceleration from A to B

b) Acceleration from B to C

Solution,

Acceleration can be calculated by dividing the change in velocity (measured in meters per second) by the time taken for the change (in seconds.)

a) v1 = 0 m/s

   v2 = 25 m/s

   t1 = 0 s

   t2 = 3 s

   $a = \frac{(v2 - v1)}{(t2 - t1)}$

   $a = \frac{(25 - 0)}{(3 - 0)}$

  a = 8.33 $m/s^{2}$

b) v1 = 25 m/s

   v2 = 17 m/s

   t1 = 3 s

   t2 = 4 s

   $a = \frac{(v2 - v1)}{(t2 - t1)}$

   $a = \frac{(17 - 25)}{(4 - 3)}$

   a = -8 $m/s^{2}$   (i.e. negative acceleration or retardation)

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