A particle follows a straight path of motion from point A to

A particle follows a straight path of motion from point A to B and C, measured along axis s. The velocity at t=0, s=0 is V_A = 2 m/s. Acceleration vs. time data is recorded and shown in figure below. Determine the velocity of the car when point B is reached and the distance travelled between points A and B. Determine the velocity of the car when point C is reached and the distance travelled between points B and C.

Solution

a) i) Velocity at point B cab be calculated by using the following equation (1) :

v = u + at                         (1)

Here , u = 2 m/sec , a = 3 m/sec² , t = 4 sec

putting values in equation (1)

v_B , velocity at point B = 2 + 3 x 4 = 14 m/sec

Hnece velocity at point B is 14 m/sec

ii ) Distance travelled between A and B ( s ) can be calculated by using the eqation (2) given below :

s = ut + 1/2 at²                                                  (2)

Here ,   u = 2 m/sec , a = 3 m/sec² , t = 4 sec

putting values in equation (2)

s = 2 x 4 + (1/2) x 3 x 4² = 8 + 24 = 32 m

Hence distance travelled between A & B = 32

b) i) Velocity at point C can be calculated by using the equation given above .

In this case since acceleration is reducing to zero from a value of 3 m/sec² , in a linear relationship. Average value of the acceleration has to be considered in the equation .

Average acceleration from B to C = ( 3 + 0 )/2 = 1.5 m/sec²

Other data u = velocity at point B = 14 m/sec , t = 4 sec

putting values in equation (1)

v_C , velocity at point C = 14 + 1.5 x 4 = 20 m/sec

Hence velocity at point C is 20 m/sec

ii) Distance travelled between B and C ( s ) can be calculated by using the eqation (2) given above

Here ,   u = 14 m/sec , a = 1.5 m/sec² , t = 4 sec

putting values in equation (2)

s = 14 x 4 + (1/2) x 1.5 x 4² = 56 + 12 = 68 m

Hence distance travelled between B & C is 68 m