Question

A,B and C are three points lying in that order on a straight line. A body is projected from B towards A with speed 3m/s. The body experiences an acceleration of 1m/s² towards C. If BC=20m, find the time taken to reach C and the distance travelled by the body from the moment of projection until it reaches C.​

Answer 1

Answer:

The total time taken to reach C = 10 s

The total distance traveled by the body = 29 m

Step-by-step explanation:

Whereby the points are vertically oriented, we have;

v = u + at

Where;

a = Acceleration of the body = -1 m/s²

u = Initial velocity = 3 m/s

v = Final velocity, = 0 m/s at maximum height

t = Time taken

0 = 3 + (-1)× t

t = 3 s

The maximum height S is given as follows;

S = u·t  + 1/2·a·t² = 3×3 + 1/2×(-1)×3² = 4.5 m

Therefore, the total distance, $S_t$, from maximum height to C = 45 + 20 = 24.5 m

The total time is given from the total distance as follows;

$S_t$  = u$_m$·t$_t$  + 1/2·a$_d$·t$_t$²

Where;

u$_m$ = Initial velocity from the maximum height = 0 m/s

a$_d$ = Downward acceleration = Opposite sign to upward acceleration = 1 m/s²

24.5  = 0·t$_t$  + 1/2·(1)·t$_t$²

49 = t$_t$²

t$_t$ = √ 49 = 7 s

The total time taken to reach C = Initial time to reach maximum height + Time from maximum height = 3 s + 7 s = 10 s

The total time taken to reach C = 10 s

The total distance traveled by the body = Distance from B to maximum height + Distance from maximum height to C

The total distance traveled by the body = 4.5 m + 24.5 = 29 m.

The total distance traveled by the body = 29 m.