Question

a dog initialy running at a speed velocity of two 2ms-2 accelerates at a constant rate of 1.5 ms-2 for 5s calculate 1 the distance covered by the day from the initial position 2 the final velocity in kmh-1​

Answer 1

Answer:

initial velocity =2m/s

acceleration = 1.5m/s2

time=10s

1) distance covered

》s=ut+1/2at^2

distance= 2×10 + 1/2× 1.5 × (5)^2

=20 +1/2 × 1.5× 25

=20+1.5×50

=38.75m

2) final velocity

final velocity= u+at

=2 + 1.5×5

= 9.5m/s