Question

An object is pushed along a rough horizontal surface and released. It slides for 11.0 s before
coming to rest and travels a distance of 22.0 cm during the last 1.5 s of its slide. Assuming the
acceleration to be uniform throughout
(a) How fast was the object travelling upon release?
Time taken to slide t = 11s
Distance traveled during 1.5s, s=22cm = 0.22m
D = V2t – (1/2) at2
2d/-t
2= a
-2(0.22) / (1.5) 2 = a
-0.196 m/s2 = a
a = (V2 – V1) / t
V2 – V1 =at
V2 – at = V1
0 – (-0.196m/s) (11) = v1
2.156 m/s = V1

(b) How fast was the object travelling when it reached the halfway position in its slide?
Time taken to slide t = 11s
V1 = 2.156 m/s
T = 11s
V2 = 0
D =?
D = v1 t + ½ at2
D = (2.156m/s) (11s) + ½ (-0.196m/s2
) (10s)2
D = 13.92m
Halfway position = 11m
V1 = 2.156 m/s
A = -0.196 m/s2
2 ad = V22 – V12
V12 + 2ad = V22
(2.156 m/s) 2 + 2 (-0.196 m/s2
) (11m) = V22
0.336m/s = V22
√0.336m/s = V2
0.58 m/s = V2

Are these correct?

Answer 1

Yes, your calculations for both parts (a) and (b) are correct.