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(i) From A to B,
Initial velocity (u) = 5m/s
Final velocity (v) = 2m/s
Time taken = from 2s to 3s= 1 sec
Acceleration = (v-u) /t = (2-5) /1 = -3 m/s²
(ii) Last 4 sec means from 3s to 7s.
Velocity from 3s to 5 s= 2 m/s
Time taken = 2 sec
Distance travelled = velocity×time
= 2m/s × 2s= 4 m
In a v-t graph distance is also calculated by finding the area under the graph.
So distance travelled from 5s to 7 s
= area of the triangle formed
=1/2 × base × height
=1/2 ×2×2= 2
Hence total distance travelled = 4m+2m= 6 m
Hope this helps!
Initial velocity (u) = 5m/s
Final velocity (v) = 2m/s
Time taken = from 2s to 3s= 1 sec
Acceleration = (v-u) /t = (2-5) /1 = -3 m/s²
(ii) Last 4 sec means from 3s to 7s.
Velocity from 3s to 5 s= 2 m/s
Time taken = 2 sec
Distance travelled = velocity×time
= 2m/s × 2s= 4 m
In a v-t graph distance is also calculated by finding the area under the graph.
So distance travelled from 5s to 7 s
= area of the triangle formed
=1/2 × base × height
=1/2 ×2×2= 2
Hence total distance travelled = 4m+2m= 6 m
Hope this helps!
Answered by
0
we have to multiply the speed into time for getting the distance so the distance equal to 1.28 into 1000 so the answer is 200
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